Question

A beaker with 2.00×10² mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 6.70 mL of a 0.450 M HCl solution to the beaker. How much will the pH change? The pKa of acetic acid is 4.740.

Express your answer numerically to two decimal places. Use a minus ( − ) sign if the pH has decreased.

Answer 1

ANSWER : Change in pH of buffer solution = - 0.269

Given

1) Volume of buffer solution = 200 ml = 0.200 L

2) pH of buffer solution = 5.000

3) pKa of acetic acid = 4.740

4) [CH₃COOH] + [CH₃COO-] = 0.10 M

First we need to calculate molar concentrations of acetic acid and acetate in the buffer solution.

We know that, pH of buffer solution is calculated by using Henderson's equation.

pH = pKa + log [CH₃COO-] / [CH₃COOH]

$\therefore$ 5.000 = 4.740 + log [CH₃COO-] / [CH₃COOH]

log [CH₃COO-] / [CH₃COOH] = 5.000 -4.740 = 0.260

[CH₃COO-] / [CH₃COOH] = 10 ^0.260 = 1.820

[CH₃COO-] = 1.820 [CH₃COOH] ---------------->(1)

We have given information that, [CH₃COOH] + [CH₃COO-] = 0.10 M

putting equation 1 into above relation, we get

$\therefore$ [CH₃COOH] + 1.820 [CH₃COOH] = 0.10

2.820  [CH₃COOH] = 0.10

[CH₃COOH] = 0.10 / 2.820 = 0.03546 M

Putting above value in equation (1) , we get

[CH₃COO-] = 1.820 ( 0.0355 ) = 0.06454 M

Now, calculate moles of acetic acid and acetate ions in 200 ml buffer solution.

We have, Molarity = No .of moles of solute / volume of solution in L

$\therefore$ No. of moles of CH₃COOH = 0.03546 mol / L ( 0.200 L ) = 0.007092 mol

No. of moles of CH₃COO^- = 0.06454 mol / L ( 0.200 L) = 0.01291 mol

No. of moles of HCl = 0.450 mol / L ( 0.0067 L) = 0.003015 mol

Consider reaction of HCl with buffer solution.

CH₃COO^- + HCl $\rightarrow$ CH₃COOH

Let's use ICE table.

Concentration (Moles)	CH3COO-	HCl	CH3COOH
Initial	0.01291	0.003015	0.007092
Change	- 0.003015	-.003015	+ 0.003015
Equilibrium	0.009895	0.000000	0.01011

After adding HCl volume of buffer is 200 ml + 6.70 ml = 206.70 ml = 0.2067 L

$\therefore$ [CH₃COO^- ] =No. of moles / volume of solution in L

= 0.009895 mol / 0.2067 L

= 0.04787 M

[CH₃COOH] = 0.01011 mol / 0.2067 L = 0.04891 M

Therefore, pH of buffer solution after addition of HCl is

pH = 4.740 + log 0.04787 / 0.04891

pH = 4.740 - 0.009334

pH = 4.731

$\therefore$ Change in pH = Final pH - Initial pH = 4.731 - 5.000 = - 0.269