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Concentration Expression
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Percentage Weight-In-Volume
The grams of solute or constituent in 100 ml of solution.
The volume, in milliliter, represents the weight in grams of solution or liquid preparation as if it were pure water.
Volume of solution (ml)( representing grams ) X % ( expressed as a decimal ) = g of solute or constituent.
Weight of solute ( g )
percentage Weight-In-Volume = ---------------------------------- X 100 Volume of solution ( ml )
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Example I
How many grams of dextrose are required to prepared
4000 ml of a 5 % solution ?
4000 ml represent 4000 gm of solution.
5 % = 0.05
g of solute or constituent = volume (ml) X % ( expressed as
a decimal ).
= 4000 g X 0.05
= 200 g
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Example II
How many grams of potassium permanganate should be used in compounding the following prescription?Rx Potassium Permanganate 0.02 % Purified Water ad 250 ml Sig. As directed
250 ml represent 250 gm of solution.0.02 % = 0.0002
g of solute or constituent = volume (ml) X % ( expressed as a decimal ). = 250 g X 0.0002 = 0.05 g
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Weight of solute ( g )
percentage Weight-In-Volume = ---------------------------------- X 100
Volume of solution ( ml )
Example
What is the percentage strength ( w/v ) of a solution of urea, if 80
ml contain 12 g ?
12
% w/v = -------- X 100 = 15 %
80
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Weight of solute ( g )
percentage Weight-In-Volume = ------------------------------------ X 100 Volume of solution ( ml )
Example How many milliliters of a 3 % w/v solution can be made from 27 g of ephedrine sulfate ?
27 3 = -------- X 100 x 27 x = --------- X 100 = 900 ml 3Volume ( in ml ) = 900 ml
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Percentage Volume-In-Volume
The milliliters of solute or constituent in 100 ml of solution.
Volume of solution (ml) X % ( expressed as a decimal ) = ml of solute or constituent.
Volume of solute ( ml )
percentage Volume-In-Volume = -------------------------------- X 100
Volume of solution ( ml )
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Example I
How many milliliters of liquide phenol be used in
compounding the following prescription?
Rx liquide phenol 2.5 %
Calamine Lotion ad 240 ml
Sig. For external use.
Volume (ml) X % ( expressed as a decimal ) = ml of solute or constituent.
240 X 0.025 = 6 ml
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Volume of solute ( ml )
percentage Volume-In-Volume = ---------------------------------- X 100
Volume of solution ( ml )
Example I
In preparing 240 ml of a certain lotion, a pharmacist used 4 ml of
liquefied phenol. What was the percentage ( v/v ) of liquefied
phenol in the lotion ?
4
% v/v = -------- X 100 = 1.67 %
240
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Volume of solute ( ml )
percentage Volume-In-Volume = ---------------------------------- X 100 Volume of solution ( ml )
Example IIWhat is the percentage strength ( v/v ) of solution of 800 g of liquid with a specific gravity of 0.800 in enough water to make 4000 ml ?
Volume = Weight / Specific Gravity. Volume = 800 / 0.8Volume = 1000 ml 1000% v/v = -------- X 100 = 25 % 4000
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Volume of solute ( ml )
percentage Volume-In-Volume = ---------------------------------- X 100 Volume of solution ( ml )
Example IIIPeppermint spirit contains 10 % of ( v/v ) of peppermint oil. What
volume of the spirit will contain 75 ml of peppermint oil ? 75 10 = -------- X 100 x
75 x = -------- X 100 = 750 ml 10
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Percentage Weight-In-Weight
The grams of solute or constituent in 100 grams of solution.
Weight of solution (g) X % ( expressed as a decimal ) = g of solute or constituent.
Weight of solute ( g )
percentage Weight-In- Weight = ------------------------------- X 100
Weight of solution ( g )
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Example I
How many grams of phenol should be used to prepare 240
g of 5 % ( w/w ) solution in water ?
Weight of solution (g) X % ( expressed as a decimal ) = g of solute or constituent.
240 X 0.05 = 12 g
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Example II
How many milligrams of hydrocortisone should be used in
compounding the following prescription?
Rx hydrocortisone 0.125 %
Hydrophilic Ointment ad 10 g
Sig. Apply.
Weight of solution (g) X % ( expressed as a decimal ) = g of solute or constituent.
10 X 0.00125 = 0.0125 g = 12.5 mg
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Example IIIHow many grams of a drug substance are required to make 120 ml of a 20 % ( w/w ) solution having a specific gravity of 1.15 ?
Volume = Weight / Specific Gravity. Weight = Volume X Specific Gravity. Weight = 120 X 1.15Weight = 138 g ( weight of 120 ml of solution )
Weight of solution (g) X % ( expressed as a decimal ) = g of solute or constituent.
138 X 0.2 = g of solute or constituent. = 27.6 g plus enough water to make 120 ml.
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Weight of solute ( g ) percentage Weight-In- Weight = ------------------------------ X 100
Weight of solution ( g )
Example IV
If 1500 g of a solution contain 75 g of a drug substance, what is
the percentage ( w/w ) of the solution ?
75
% w/w = ---------- X 100 = 5 %
1500
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Weight of solute ( g ) percentage Weight-In- Weight = ------------------------------ X 100
Weight of solution ( g )
Example VIf 5 g of boric acid are add to 100 ml of water, what is the percentage strength ( w/w ) of the solution ?
100 ml of water weight 100 g100 g + 5 g = 105 g, weight of the solution 5 % w/w = ---------- X 100 = 4.76 % 105
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Milligram Percent
The number of milligrams of substance in 100 ml of liquid.
It is used frequently to denote the concentration of a drug or natural substance in a biologic fluid, as in the blood.
The statement that the concentration of non-protein nitrogen in the blood is 30 % means that each 100 ml of blood contains 30 mg of non-protein nitrogen.
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Example I If a patient is determined to have a serum cholesterol level of 200 mg/dl (a) What is the equivalent value expressed in terms of milligrams percent?
(a) 200 mg/dl = 200 mg / 100 ml = 200 mg%.
(b) How many milligrams of cholesterol would be present in 10 ml sample of the patient's serum ?
(b )200 (mg) 100 ml
x (mg) 10 mlx ( mg ) = 200 X 10 / 100 = 20 mg.
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Example II If a patient is determined to have a serum cholesterol level
of 200 mg/dl, what is the equivalent value expressed in term of millimoles ( mmol ) per liter ?
Molecular weight of cholesterol = 387.
1 mol cholesterol = 387 g.
1mmol cholesterol = 387 mg.
200 mg/dl = 2000 mg/L.
387 ( mg ) 1 (millimoles )
2000 ( mg ) x (millimoles )
Therefore x = ( 1 X 2000 ) / 387
= 5.17 mmol / L
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Part Per Million ( PPM ) & Part Per Billion ( PPB )
The strength of very dilution solution are commonly expressed in terms of part per million ( PPM ) & part per billion ( PPB ), i.e. the number of parts of the agent per 1 million or 1 billion parts of the whole.
For example, fluoridated drinking water (used to reduce dental caries) often contains 1 part of fluoride per million parts of drinking water.
The ppm or ppb concentration of a substance may be expressed in quantitatively equivalent value of percent strength or ratio strength.
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Example I Express 5 ppm of ion in water in percent strength and ratio
strength ?
5 ppm = 5 parts in 1,000,000 parts
= 1 : 200,000 ( ratio strength )
= ( 1 / 200,000 ) X 100
= 0.0005 % ( percent strength )
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Example II
The concentration of a drug additive in an animal feed is 12.5 ppm. How many milligrams of the drug should be used in preparing 5.2 kg of feed ?
12.5 ppm = 12.5 g ( drug ) in 1,000,000 g ( feed )
Thus
12.5 g ( drug ) 1,000,000 g ( feed )
x g ( drug ) 5,200 g ( feed )
x = ( 5,200 X 12.5 ) / 1,000,000
= 0.065 g
= 65 mg
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Molarity
Concept of mole
Atoms & molecules are too small, tiny to weight so the concept of mole which is theoretical value or unit expressing the formula weight of substance in grams.
e.g.
gm atom of O2 M.Wt is 16 = 16 gm = 1 molgm ion of Cl- M.Wt is 35.5 = 35.5 gm = 1 molgm molecules of H2O M.Wt is 18 = 18 gm = 1 mol
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Molar Concentration
Moles / Liter is one of the most useful units for describing concentration.
Molarity ( M ) is the number of moles of solute / Liter of solution ( not solvent ).
M X L = moles.
M X L = Weight in grams / molecular weight.
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Example I How can prepare 500 ml of 0.15 M Na2CO3 solution ?
i.e. How many grams needed ?
M X L = Weight in grams / molecular weight.
0.15 X 0.5 = Weight in grams / 106
Weight in grams = 0.15 X 0.5 X 106
Weight in grams = 7.95 g
Weighing 7.95 g of Na2CO3 & placed in volumetric flask 500 ml & diluted to 500 ml .
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Example II What is the molarity of NaCl solution 10 g / 100 ml ?
M X L = Weight in grams / molecular weight. M X 0.1 = 10 / 58.5.
M = 1.71 M
Example II
What is the molar concentration of 1 % (w/v) NaCl solution?
1 % = 1 g of NaCl in 100 ml
M X L = Weight in grams / molecular weight. M X 0.1 = 1 / 58.5.
M = 0.171 M
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Example IV Describe the preparation of 2 L of 0.2 M HCl solution starting
with a concentrated HCl solution ( 28 % w/w, specific gravity = 1.15 ) ?
M X L = Weight in grams / molecular weight.0.2 X 2 = x / 36.5.Weight in grams = 0.2 X 2 X 36.5Weight in grams = 14.6 g ( need to prepare the solution )
The stock of HCl is not pure only 28 % HCl: 28 g HCl 100 g of solution.
14.6 g HCl x g of solution.x = ( 14.6 X 100 ) / 28x = 52.143 g
52.143 g should be taken from a solution to take it in ml instead of grams ( specific gravity = weight / volume )Volume = weight / specific gravity = 52.143 / 1.15 = 45.34 ml
45.34 ml placed in a volumetric flask and completed to 2 liters.
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Molality ( m )
number of moles of solute / kg of solvent ( not solution ).
m X kg ( solvent ) = moles.
m X kg ( solvent ) = Weight in grams / molecular weight.
Molality is useful in describing the ratio of moles of solute to solvent.
The value of ( m ) does not change with temperature but that of ( M ) dose.
We would not even have to use a volumetric flask, because we weight the solvent.
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Example I Calculate the molality HCl solution ( 28 % w/w ) ?
m X kg ( solvent ) = Weight in grams / molecular weight.
Weight of solvent = 100 – 28 = 72 g ( from 28 % )
= 0.072 kg
m X kg ( solvent ) = Weight in grams / molecular weight.
m X 0.072 = 28 / 36.5
m = 28 / ( 36.5 X 0.072 )
m = 10.65 ( m)
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Example II
Calculate the molal concentration of 1 % ( w/v ) NaCl solution ?
Where : molecular weight = 58.5 & specific gravity = 1.0055
1 % mean 1 g NaCl in 100 ml solution
Weight of solution = volume X specific gravity
= 100 X 1.0055 = 100.55 g
Weight of solvent = weight of solution – weight of solute
= 100.55 – 1 = 99.55 g
m X kg ( solvent ) = Weight in grams / molecular weight.
m X 0.09955 = 1 / 58.5
m = 1 / ( 58.5 X 0.09955 )
m = 1.72 ( m)
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Example III To prepare 0.15 m solution of NaCl, to prepare this solution with
this conc, how many grams of NaCl should be dissolved in 500 gm of water ?
m X kg ( solvent ) = Weight in grams / molecular weight.
0.15 X 0.5 = Weight in grams / 58.5Weight in grams = 0.15 X 0.5 X 58.5Weight in grams = 4.39 g
Thus, 4.39 g of NaCl is dissolved in 500 gm of H2O to give the desired concentration .
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Mole Fraction
Ratio of moles of one constituent solute or solvent of solution to the total number of moles of all constituents ( solute & solvent ).
y1 = n1/(n1+n2) y2 = n2 /(n1+n2) Where
n1,n2 are the number of moles.
y1 is the mol fraction of solute.
y2 is the mol fraction of solvent.
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Example IWhat is the mole fraction of both constituent in 1 % ( w/v ) NaCl
solution ( specific gravity = 1.0053 ) and what is the molality ?
Volume = Weight / Specific Gravity. Weight = Volume X Specific Gravity. Weight = 100 X 1.0053Weight = 100.53 g solution1 g NaCl in ( 100 X 1.0053 = 100.53 g solution )Solvent 100.53 – 1 = 99.53 g solvent.
m X kg = Weight in grams / molecular weight.m X 99.53/1000 = 1 / 58.5.
m = 1 / ( 58.5 X 0.09953)m = 0.172 ( m )
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Cont. Example I
What is the mole fraction of both constituent in 1 % ( w/v ) NaCl solution ( specific gravity = 1.0053 ) and what is the molality ?
No. of moles = Weight in grams / molecular weight
No. of moles of solute = 1 / 58.5 = 0.171
No. of moles of solvent = 99.53 / 18 = 5.529
y1 = n1/(n1+n2) ) y2 = n2 /(n1+n2)
Mole fraction of solute = 0.171 / ( 0.171 + 5.529 ) = 0.003
Mole fraction of solvent = 5.529 / ( 0.171 + 5.529 ) = 0.996
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Mole Percent
Mole Percent = mole fraction X 100
In the last example :
Mole fraction of solute = 0.171 / ( 0.171 + 5.529 ) = 0.003Mole Percent = mole fraction X 100Mole Percent = 0.003 X 100 = 0.3 %
Mole fraction of solvent = 5.529 / ( 0.171 + 5.529 ) = 0.996Mole Percent = mole fraction X 100Mole Percent = 0.996 X 100 = 99.6 %
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Normality ( N )
Equivalent weight is the weight in grams of one equivalent or the quantities of substance that combine with 1.008 gm H+.
The EQUIVALENT WEIGHT is the amount of solute needed to be
the equivalent of one mole of hydrogen ions. Therefore, the equivalent weight is dependent on the valence of the solute. For solutes with a valence of one (i.e. NaCl) the molecular weight and equivalent weight are the same. When the valence of the solute is more than one (i.e. H3PO4, valence = 3), then the equivalent weight is equal to the molecular weight divided by the valence.
The equivalent weight ( one equivalent ) of an acid or base is that contains 1 g atom.
Equivalent weight = molecular weight / n.n = number of replaceable H or OH for acid or base.n = number of electrons lost or gained in oxidation, reduction
reaction.
Equivalent weight = atomic weight / number of equivalent per atomic weight.
e.g. F & O2
1 equivalent of F = molecular weight = 19
1 equivalent of O2 = molecular weight / Valency = 16 / 2 = 838
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Normality ( N ) = grams of equivalent weight of solute / Liter of solution ( not solvent )
L X N = equivalent
N = weight in gram / (equivalent weight x volume)
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Concentration Expression: ExpressionSymbolDefinition
%Weight in Volume%W/VGrams of solute in 100 ml of solution.
%Volume in Volume%V/VMilliliters of solute in 100 ml of solution.
%Weight in Weight %W/WGrams of solute in 100 g of solution.
Milligram% Milligrams of solute in 100 ml of solution.
part per million or billionPPM,PPBthe number of parts of the agent per 1 million or 1 billion parts of the whole.
Molarity M, CMole ( gram molecular weights) of the solute in 1 liter of solution.
MolalityMMole of the solute in 1000 g of solvent.
Mole fractionX, NRatio of the moles of one constituent (e.g. the solute ) of a solution to the total moles of all constituents ( solute & solvent ).
Mole percentMoles of one constituent in 100 moles of the solution ( by multiplying mole fraction by 100 ).
NormalityNGram equivalent weight of solute in 1 liter of solution.
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Buffer
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pH
pH = - log [ H+] or pH = - log [ H3O+ ]
Example I
What is the pH of solution with [ H+] = 32 X 10-5 M/L ?
pH = - log [ H+]
pH = - log 32 X 10-5
pH = 3.495
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Example II
The hydronium ion concentration of 0.1 M solution was found to the 1.32 X 10-3 M, What is the pH of the solution?
pH = - log [ H3O+ ]
pH = - log [ 1.32 X 10-3 ]
pH = 2.88
Example III
If the pH of a solution is 4.72, what is the hydrogen ion concentration?
pH = - log [ H+ ]
- log [ H+ ] = 4.72
log [ H+ ] = - 4.72 take anti-log for both side
[ H+ ] = 1.91 X 10-5
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pH = 0 H+ = 1 M/l = 1 X 100 = 1 pH = 1 H+ = 1 X 10-1 = 0.1 M/L pH = 2 H+ = 1 X 10-2 = 0.01 M/L pH = 3 H+ = 1 X 10-3 = 0.001 M/L
Note : the change in one pH unit means 10 fold change in [H+].
pH + pOH = 14
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Introduction When a minute trace of hydrochloric acid is added to pure
water, a significant increase in hydrogen-ion concentration occur immediately.
In a similar manner, when a minute trace of sodium hydroxide is added to pure water, it cause correspondingly large increase in the hydroxyl-ion concentration.
These change take place because water alone cannot neutralize even trace of acid or base, i.e. it has no ability to resist change in hydrogen-ion concentration or pH.
Therefore, it is said to be unbuffered. E.g. CO2 + H2O H2CO3 decrease pH from 7 to 5.8
These change of pH are of great concern in pharmaceutical
preparation also NaCl solution ability to resist change of pH.
To ensure stability and solubility, we used to control pH by using
a buffer
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Buffer :Compound or a mixture of compound which by presence in
solution to resist change in pH up of addition of small quantities of acid or base or a solvent.
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Adjustment of pH by the buffer
pH important for stability & solubility, therefore should be adjusted pH by buffer .
Also pH play important role in :
1-Parenteral dosage form.2-Eye drops.
3-Nasal drops.
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Adjustment of pH by the buffer
Degree of acidity and alkalinity depends on the relative concentration of H+ ion and OH- ion.
if H+ > OH- = acidic
H+ = OH- = neutral
H+ < OH- = alkaline
Acidity and alkali may be strong or weak:
1- weak acid, pH 3.5- 7 2- strong acid, pH 0-3.53- weak base, pH 7-10 4- strong base, pH 10.5-14
0 3.5 7 10.5 14
Strong acidWeak acidWeak baseStrong base
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The product of H+ ion and OH- ion in the any aqueous liquid is constant i.e. Kw = [ H+] [ OH- ]
Increase of one tend to decrease of another.
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Some notes about buffer:
1-Buffers solution should be prepared using freshly boiled and cooled water.
2- Buffers solution should be stored in containers of Alkali-free glass.
3-Buffers solution should be discarded no later than three months from the date of manufacture.
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Selection of buffer system depends on:
1. pH rang.
2. Buffer capacity desired.
3. The purpose for which it is required.
4. Compatibility with active ingredients.
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Method of preparation of buffer
1. Buffer equation.
2. Buffer table.
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Buffer solution consist of mixture of weak acid and its conjugate base or weak base and its conjugate acid.
Weak acid + its salt
e.g. acetic acid + sodium acetate
CH3COOH CH3COONa
Weak base + its salt
e.g. Ammonia + Ammonium chloride
NH3 + NH4Cl
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Henderson-Hasselbalch Equationbuffer equation for weak acid and its salt
Dissociation Constant of weak acid is given by the equation :
Where :
A- = Salt
HA = Acid
we isolate the H+ and put it on the left-hand side of the equation:
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take the negative log of each of the three terms in the last
equation, they become:
- log [H+] this is the pH
- log Ka this is the pKa
- log ([HA] / [A-]) to get rid of the negative sign + log ([A-] / [HA])
Inserting these last three items (the pH, the pKa and the
rearranged log term), we arrive at the Henderson-Hasselbalch
Equation:
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common way the Henderson-Hasselbalch Equation is presented
in a textbook explanation:
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Remember that, in a buffer, the two substances differ by only a proton. The substance with the proton is the acid and the substance without the proton is the salt.
However, remember that the salt of a weak acid is a base (and the salt of a weak base is an acid).
Consequently, another common way to write the Henderson-Hasselbalch Equation is to substitute "base" for "salt form" (sometimes you will see "conjugate base" or "base form"). This is probably the most useful way to decribe the interactions between the acidic form (the HA) and the basic form (the A-).
Here it is:
Remember this: the base is the one WITHOUT the proton and the acid is the one WITH the proton.
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Buffer equation of weak base & its salt
Dissociation Constant of weak acid is given by the equation :
( B + ) ( OH- )
Kb = --------------------
( BOH)
Where : B + = Salt
BOH = Base
[Base]
pH = pKw – pKb + log ----------
[ Salt]
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An alternate form of the Henderson-Hasselbalch Equation
The last discussion used pH and pKa. There is a alternate form of the Henderson-Hasselbalch Equation using pOH and pKb.
[Salt]
pOH = pKb + log -----------
[Base]
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Mechanism of action of buffer
1 – Acid / Its salt e.g. CH3COOH/CH3COONa.
CH3COOH + OH- CH3COO- + H2O
This means acid will react with base (OH- ( to neutralize it.
CH3COO-/Na+ + H3O+ CH3COOH + H2O
This means salt of weak acid will react with acid ( H3O ) to neutralize it.
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Mechanism of action of buffer… cont
2 – Base / Its salt e.g. NH3/NH4Cl.
NH3 + H3O+ NH4+- + H2O
This means base will react with acid to neutralize it.
NH4+ + OH- NH3 + H2O
This means salt of weak base will react with base (OH- ) to neutralize it.
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pH + pOH = 14
For acid & its salt :
[Salt]pH = pKa + log ---------- [Acid]
For base & its salt :
[Salt]pOH = pKb + log ----------- or [Base]
[Base]pH = pKw – pKb + log ---------- [ Salt]
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Example I What is the molar ratio of salt/acid required to prepare an acetate
buffer pH = 5 ?
Ka = 1.8 X 10-5
pKa = - log Ka
pKa = - log 1.8 X 10-5 = 4.75
pH = pKa + log salt/acid
5 = 4.75 + log salt/acid
log salt/acid = 5 – 4.75 = 0.25 take anti-log for both side
salt/acid = 1.78
So, Ratio of salt/acid = 1.78/1
Mole fraction of acid = 1 / ( 1.78 + 1) =0.3597 X 100 = 35.97 %
Mole fraction of salt = 1.78 / ( 1.78 + 1) =0.6403 X 100 = 64.03 %
mole fraction multiply by 100 to get of mole %
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Example IIPrepare 200 ml of acetate buffer pH = 6 with molar conc. = 0.4 M
& Ka = 1.8 X 10-5 ?
pKa = - log Ka
pKa = - log 1.8 X 10-5 = 4.75pH = pKa + log salt/acid6 = 4.75 + log salt/acidlog salt/acid = 6 – 4.75 = 1.25 take anti-log for both sidesalt/acid = 17.78/1
Molecular weight of acetic acid ( CH3COOH ) = 60Molecular weight of sodium acetate ( CH3COONa ) = 82
Weight of acid = mole fraction X Conc. X M.wt X V (L) = 1/18.78 X 0.4 X 60 X0.2 = 0.26 gWeight of salt = mole fraction X Conc. X M.wt X V (L) = 17.78/18.78 X 0.4 X 82 X0.2 = 6.21 g
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Example III If we add 0.1 M sodium acetate to 0.09 M acetic acid .What is the
pH if you know Ka = 1.8 X 10-5 ?
pKa = - log Ka
pKa = - log 1.8 X 10-5 = 4.75
pH = pKa + log salt/acid
pH = 4.75 + log 0.1/0.09
pH = 4.796
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Example IV
What is the pH of of a solution containing 0.1 mole of ephedrine base and 0.01 mole of ephedrine HCl / liter of solution ?pKb of ephedrine = 4.64
pH = pKw - pKa + log base/salt
pH = 14 - 4.64 + log 0.1/0.01
pH = 9.36 + log 10 = 10.36
Or
pOH = pKb + log salt/base
pOH = 4.64 + log 0.01/0.1
pOH = 4.64 + log 0.1 = 3.64
pH + pOH = 14
pH = 14 – pOH
pH = 14 – 3.64 = 10.36
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Change in pH with addition of an acid or base
Calculate the change in pH of a buffer solution with the addition of a given amount of acid or base in the following example :
Example
Calculation the change in pH after adding 0.04 mol of sodium hydroxide to a liter of a buffer solution containing 0.2 M conc. of sodium acetate and acetic acid. The pKa value of acetic acid is 4.76 at 25 OC.
The pH of the buffer solution is calculated by using the buffer equation
pH = pKa + log salt/acid
pH = 4.76 + log 0.2/0.2
pH = 4.76
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The addition of 0.04 mol of sodium hydroxide converts 0.04 mol of acetic acid to 0.04 mol of sodium acetate. Consequently, the conc. Of the acetic acid is decrease and the conc. Of the sodium acetate is increase by equal amounts, according to the following equation :
salt + base
pH = pKa + log ------------------------
acid - base
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salt + base
pH = pKa + log -----------------------
acid - base
0.2 + 0.04
pH = 4.76 + log -------------------
0.2 - 0.04
pH = 4.76 + 0.1761 = 4.9361 = 4.94
Because the pH before addition sodium hydroxide was 4.74, the change in pH = 4.94 – 4.76 = 0.18 unit.
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Buffer Capacity
The ability of buffer solution to resist change in pH upon addition of acid or base.
Ka . [H+]
Buffer Capacity “ B “ = 2.303 X C X -----------------
( Ka + [H+] )2
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Example I
At hydrogen ion conc. of 1.75 X 10-5 (4.76), what is the capacity of a buffer containing 0.1 mole each of acetic acid and sodium acetate / liter of solution ?
Total C = [acid] + [ salt] = 0.1 + 0.1 = 0.2 mol/L
Ka . [H+]Buffer Capacity “ B “ = 2.303 X C X -----------------
( Ka + [H+] )2
(1.75 X 10-5 ) . (1.75 X 10-5 )Buffer Capacity “ B “ = 2.303 X 0.2 X ---------------------------------------- [(1.75 X 10-5 ) + (1.75 X 10-5 )]2
B = 0.115
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Thank you