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 )

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

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

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 ?

% w/v = -------- X 100 = 15 %

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

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

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

percentage Volume-In-Volume = ---------------------------------- X 100

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 ?

% v/v = -------- X 100 = 1.67 %

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

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

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.

percentage Weight-In- Weight = ------------------------------- X 100

Weight of solution ( g )

Example I

How many grams of phenol should be used to prepare 240

g of 5 % ( w/w ) solution in water ?

240 X 0.05 = 12 g

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.

10 X 0.00125 = 0.0125 g = 12.5 mg

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 )

138 X 0.2 = g of solute or constituent. = 27.6 g plus enough water to make 120 ml.

Weight of solute ( g ) percentage Weight-In- Weight = ------------------------------ X 100

Example IV

If 1500 g of a solution contain 75 g of a drug substance, what is

the percentage ( w/w ) of the solution ?

% w/w = ---------- X 100 = 5 %

Weight of solute ( g ) percentage Weight-In- Weight = ------------------------------ X 100

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

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.

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.

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

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.

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 )

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 )

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

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.

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

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.

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 .

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

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.

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.

Example I Calculate the molality HCl solution ( 28 % w/w ) ?

Weight of solvent = 100 – 28 = 72 g ( from 28 % )

= 0.072 kg

m X 0.072 = 28 / 36.5

m = 28 / ( 36.5 X 0.072 )

m = 10.65 ( m)

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 0.09955 = 1 / 58.5

m = 1 / ( 58.5 X 0.09955 )

m = 1.72 ( m)

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 ?

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 .

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.

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 )

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

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 %

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

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)

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.

Buffer

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

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

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

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

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.

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.

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

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.

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.

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.

Method of preparation of buffer

1. Buffer equation.

2. Buffer table.

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

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:

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:

common way the Henderson-Hasselbalch Equation is presented

in a textbook explanation:

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.

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]

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]

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.

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.

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]

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 %

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

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 = 4.75 + log 0.1/0.09

pH = 4.796

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

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

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 = 4.76 + log 0.2/0.2

pH = 4.76

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

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.

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

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

Thank you

Concentration Expression

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