Question

Consider the following solutions:
Solution 1: 0.100 L of 0.25 M NaCH₃CO₂ + 0.050 L of 0.25 M HCl
Solution 2: 0.100 L of 0.25 M HCH₃CO₂ + 0.050 L of 0.25 M NaOH

1. How many moles of acid does it take to change the pH of the original Solution 1 by 0.30 pH units?

2. How many moles of acid does it take to change the pH of the original Solution 2 by 0.30 pH units?

Answer 1

Solution 1: 0.100 L of 0.25 M 0.050 L of 0.25 M HCl

NaCH3CO2 +

Number of moles of

HC1 = 0.050 L x 0.25 mol/L = 0.0125 mol

Number of moles of

NaCH3C02 = 0.100 L x 0.25 mol/L = 0.025 mol

0.0125 moles of HCl combine with 0.0125 moles of
NaCH3CO
to form 0.0125 moles of
HCHCO
.

NaCH3CO, + HCl + HCH3CO, + NaCl

remains.

0.025 mol – 0.0125 mol = 0.0125 mol NaCH3CO

This is an acid buffer solution and equal amounts of $\large HCH_3CO_2 \text { and }NaCH_3CO_2$ are present.

In such cases,

pH = pk.

For acetic acid,

pka = 4.74

Hence,

pH = 4.74

Let x moles of acid are added to change pH by 0.30 pH units

New pH will be

4.74-0.30 = 4.44

conjugate base pH = pka + log10- acid] conjugate base 4.44 = 4.74 + logio- (acid] 100. (conjugate base] = 4.44 – 4.74 acid conjugate base logio = -0.30 log10 acid

conjugate base] L = 10-0.30 acid] conjugate base = 0.5012...... (1) acid

x moles of acid are added to change pH by 0.30 pH units

Number of moles of acid are

(0.0125 + ) mol

Number of moles of conjugate base are

(0.0125 – ) mol

conjugate base acid number of moles of conjugate base number of moles of acid (0.0125 - rmol (0.0125 + x) mol 0.0125 - 0.0125+

But (1)=(2)

0.0125 – = 0.5012 0.0125 + x 0.0125 - x = 0.5012(0.0125 + r) 0.0125 – I = 0.0063+ 0.50122 0.0125 - 0.0063 = 1 +0.5012.

0.0062 = 1.50120 0.0062 = 1.5012 x = 0.00415

Hence, 0.00415 moles of acid are needed.

Solution 2: 0.100 L of 0.25 M 0.050 L of 0.25 M NaOH

HCHCO, +

Number of moles of

NaOH = 0.050 L x 0.25 mol/L = 0.0125 mol

Number of moles of $\large HCH_3CO_2 = 0.100 \ L \times 0.25 \ mol/L=0.025 \ mol$

0.0125 moles of NaOH combine with 0.0125 moles of
HCHCO
to form 0.0125 moles of
NaCH3CO
.

HCHECO, + NaOH + NaCH3CO, + H.O

remains.

0.025 mol – 0.0125 mol = 0.0125 mol HCH.CO

This is an acid buffer solution and equal amounts of $\large HCH_3CO_2 \text { and }NaCH_3CO_2$ are present.

In such cases,

pH = pk.

For acetic acid,

pka = 4.74

Hence,

pH = 4.74

Let x moles of acid are added to change pH by 0.30 pH units

New pH will be

4.74-0.30 = 4.44

conjugate base pH = pka + log10- acid] conjugate base 4.44 = 4.74 + logio- (acid] 100. (conjugate base] = 4.44 – 4.74 acid conjugate base logio = -0.30 log10 acid

conjugate base] L = 10-0.30 acid] conjugate base = 0.5012...... (1) acid

x moles of acid are added to change pH by 0.30 pH units

Number of moles of acid are

(0.0125 + ) mol

Number of moles of conjugate base are

(0.0125 – ) mol

conjugate base acid number of moles of conjugate base number of moles of acid (0.0125 - rmol (0.0125 + x) mol 0.0125 - 0.0125+

But (1)=(2)

0.0125 – = 0.5012 0.0125 + x 0.0125 - x = 0.5012(0.0125 + r) 0.0125 – I = 0.0063+ 0.50122 0.0125 - 0.0063 = 1 +0.5012.

0.0062 = 1.50120 0.0062 = 1.5012 x = 0.00415

Hence, 0.00415 moles of acid are needed.