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Class Monday, Oct 11, 2004

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Another pH Buffer Problem

What is the pH of a solution prepared by mixing together100 mL of 0.2500 M ammonia and 200 mL of 0.1500 Mammonium chloride. The Kb for ammonia is 1.75 105.

Answer: pH = 9.16

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Another pH Buffer Problem

What is the pH of a solution prepared by mixing together100 mL of 0.2500 M ammonia and 200 mL of 0.1500 Mammonium chloride. The Kb for ammonia is 1.75 105.• The pKb = -log 10(1.75 x 105) = 4.757

• pKa + pKb = 14.0, pKa for the acid form (NH4+) = 9.243

• The total volume = 200 + 100 = 300 mL

• Using the H-H equation, pH = pKa + log 10(base/acid)

• pH = 9.243 + log 10{(100 x 0.2500/300) ÷ (200 x 0.1500/300)}

• pH = 9.243 + (0.0792) – 9.163 = 9.16

Note that the (no. mol of acid) > (no. mol of base); the pH will lie to the side of the pKa of which ever one is the larger, here acidic side of9.243

Answer: pH = 9.16

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Class Monday, Oct 11, 2004

pH = pKa + log 10{[base] / [acid]}

Generally, the pH range that the buffer will work most effectively is

pH = pKa ± 1.00

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Another Buffer Problem – Choose a system (conjugate pair) to make a buffer whose pH = 7.0

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Another Buffer Problem – Choose a system (conjugate pair) to make a buffer whose pH = 7.0

As mentioned in a previous slide, the pH of the buffer is roughly equal to pKa of the weak acid. From Appendix B, pages 540ff there are several system whose pKa values are close to 7.0; I am going to choose the phosphate buffer with pKa = 7.199. 

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Another Buffer Problem – Choose a system (conjugate pair) to make a buffer whose pH = 7.0

The ratio of ([HPO42-] / [H2PO4-]) is calculated from

the Henderson-Hasselbalch expression. • pH = pKa + log 10{[HPO4 2-] / [H2PO4 -]}• 7.00 = 7.199 + log 10{[HPO4 2-] / [H2PO4 -]}• log 10{[HPO4

2-] / [H2PO4-]}= 7.00 – 7.199 = - 0.199

• {[HPO4 2-] / [H2PO4 -]}= 10 -0.199 = 0.632

• This means that the ratio of {[base] / [acid]} must be 0.632:1 to have a buffer with a pH of 7.00

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Another Buffer Problem – Choose a system (conjugate pair) to make a buffer whose pH = 7.0

The ratio of ([HPO4 2-] / [H2PO4

-]) is calculated from the Henderson-Hasselbalch expression.

pH = pKa + log 10{[HPO4 2-] / [H2PO4

-]} 7.00 = 7.199 + log 10{[HPO4

2-] / [H2PO4 -]}

log 10{[HPO4 2-] / [H2PO4

-]}= 7.00 – 7.199 = - 0.199 {[HPO4

2-] / [H2PO4 -]}= 10 -0.199 = 0.632

This means that any ratio of {[base] / [acid]} = 0.632 will have a pH of 7.00

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Buffer Capacity

Buffer capacity measures the resistance the buffer solution has to changes in pH whenever an acid or a base is added. It is technically defined as the number of moles of acid or base one liter of the buffer solution can absorb with a change of pH not to exceed 1 pH unit.

The greater the concentrations of the acid and base forms, the greater is the buffer capacity. The buffer capacity is also greatest near the pKa of the acid form of the system.

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Buffer Capacity

Buffer capacity () is the number of moles of OH–or H + that 1.00 Liter of a buffer can absorb without the pH change exceeding 1 pH unit.

The buffer capacity depends on the concentrations ofthe weak acid and its conjugate base.

For the addition of base: nOH- = nHB originally presentFor the addition of acid: nH+ = nB- originally present In practice, pH starts to change drastically as nHB ornB→ 0, as is shown in the next slide.

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Buffer Capacity

The effect of adding increments of H+ or OH to a buffer system of HA and A whose pKa = 5.0 and the total concentration of = 1 M.

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Buffer Capacity

Whenever a strong acid or a strong base is added to a

buffer the following reactions occur:

1. Addition of strong base (OH-)

HB(aq) + OH-(aq) H2O + B- (aq)

2. Addition of strong acid (H+ or H3O+)

B-(aq) + H+(aq) HB(aq)

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Buffer Capacity

So long as the system has plenty of HB and B– to

consume the H+ or OH- ions that have been added

there is not a drastic change in the pH. The actual

pH will depend on the ratio of the base form : acid

form as shown in the Henderson-Hasselbalch

equation.

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Buffer Capacity Problem

What is the new pH whenever 0.100 mol of HCl is added to 1 liter of the pH 7.0 phosphate buffer chosen earlier if the [H2PO4

-] = 0.800 M.• Earlier we calculated that Base : Acid ratio needed to be 0.632, so if the [acid] = 0.800 M, the [base] = 0.632 x 0.800M = 0.506M• The addition of 0.100 mol of HCl (H+) will cause H2PO4

- to increase by 0.100 mol and the HPO4

-2 to decrease by 0.100 mol; the reaction is

HPO4-2 + H+ → H2PO4

-

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What is the new pH whenever 0.100 mol of HCl is added to 1 liter of the pH 7.0 phosphate buffer chosen earlier if the [H2PO4

-] = 0.800 M.

The new mol of HPO4-2 = (1.00L)(0.506) – 0.100

= 0.406 mol; since in 1.00 L, [HPO4-2] = 0.406 M

The new mol of H2PO4

- = (1.00)(0.800) + 0.100 = 0.900 mol; since in 1.00 L, [H2PO4

-] = 0.900M

The new pH is found by substituting the new concentration values into the H-H equation:

pH = pKa + log 10{[base] / [acid]}pH = 7.199 + log 10{0.406 / 0.900}pH = 7.199 + (- 0.346) = 6.853

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Buffer Solutions

Note that the addition of strong acid causes the pH of the buffer to become more acidic (lower pH). Conversely, the addition of a strong base would cause the pH of the buffer to become more basic (higher pH).

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Buffer Solutions

The Buffer capacity of the 0.500 M lactic acid/lactate buffer. Note the middle of the buffer range occurring at pH of = 3.85, the pKa of this system.