Question

500.0 mL of 0.130 M NaOH is added to 555 mL of 0.200 M weak acid (Ka = 7.11 × 10-5). What is the pH of the resulting buffer?

Answer 1

Concepts and reason

The concept used to solve this problem is based on the chemical equilibrium.

The pH of solution is used to determine acidity and basicity of a solution.

Fundamentals

The pH of solution can be determine by Henderson-Hasselbalch equation as follow.

${\rm{pH}} = {\rm{p}}{{\rm{K}}_{\rm{a}}} + \log \frac{{\left[ {{\rm{conjugate base}}} \right]}}{{\left[ {{\rm{acid}}} \right]}}$

Here ${\rm{p}}{{\rm{K}}_{\rm{a}}}$ is written as follow.

${\rm{p}}{{\rm{K}}_{\rm{a}}} = - \log {{\rm{K}}_{\rm{a}}}$

Here, ${{\rm{K}}_{\rm{a}}}$ is acid dissociation constant.

First calculate the number of moles by using formula as follow.

$n = \left( V \right)\left( m \right)$

Here V is volume of solution and m is molarity of the acid or base.

Now for initial moles of weak acid $\left( {{n_{{\rm{HA}}}}} \right)$ , substitute 555 mL for V and 0.200 M for m.

${n_{{\rm{HA}}}} = \left( {555{\rm{ mL}}} \right)\left( {0.200{\rm{ M}}} \right)$

Convert volume in liters.

$\begin{array}{c}\\{n_{{\rm{HA}}}} = \left( {555{\rm{ mL}}} \right)\left( {\frac{{1{\rm{ L}}}}{{1000{\rm{ mL}}}}} \right)\left( {0.200{\rm{ mol }}{{\rm{L}}^{ - 1}}} \right)\\\\ = 0.111{\rm{ mol}}\\\end{array}$

Similarly for ${n_{{\rm{NaOH}}}}$ , substitute 500 mL for V and 0.130 M for m.

${n_{{\rm{NaOH}}}} = \left( {500{\rm{ mL}}} \right)\left( {0.130{\rm{ M}}} \right)$

Convert volume in liters.

$\begin{array}{c}\\{n_{{\rm{NaOH}}}} = \left( {500{\rm{ mL}}} \right)\left( {\frac{{1{\rm{ L}}}}{{1000{\rm{ mL}}}}} \right)\left( {0.132{\rm{ mol }}{{\rm{L}}^{ - 1}}} \right)\\\\ = 0.065{\rm{ mol}}\\\end{array}$

Write the balanced chemical equation between weak acid and base as follow.

${\rm{NaOH}} + {\rm{HA}} \to {\rm{N}}{{\rm{a}}^ + } + {{\rm{A}}^ - } + {{\rm{H}}_2}{\rm{O}}$

Now, number of moles of weak acid left adding the base is calculated by subtracting the initial moles of weak acid and moles of base added.

${n_{{\rm{H}}{{\rm{A}}_{{\rm{left}}}}}} = {n_{{\rm{HA}}}} - {n_{{\rm{NaOH}}}}$

Substitute 0.111 mol for ${n_{{\rm{HA}}}}$ and 0.065 mol for ${n_{{\rm{NaOH}}}}$.

$\begin{array}{c}\\{n_{{\rm{H}}{{\rm{A}}_{{\rm{left}}}}}} = 0.111{\rm{ mol}} - 0.065{\rm{ mol}}\\\\{\rm{ = 0}}{\rm{.046 mol}}\\\end{array}$

Now, moles of ${{\rm{A}}^ - }$ formed is equal to moles of NaOH added in solution which is equal to 0.0065 mol.

Now, ${\rm{p}}{{\rm{K}}_{\rm{a}}}$of solution can be calculated by formula as follow.

${\rm{p}}{{\rm{K}}_{\rm{a}}} = - \log {{\rm{K}}_{\rm{a}}}$

Substitute $7.11 \times {10^{ - 5}}$for ${{\rm{K}}_{\rm{a}}}$.

$\begin{array}{c}\\{\rm{p}}{{\rm{K}}_{\rm{a}}} = - \log \left( {7.11 \times {{10}^{ - 5}}} \right)\\\\ = 4.148\\\end{array}$

Now, Henderson-Hasselbalch equation for this reaction is written as follow.

${\rm{pH}} = {\rm{p}}{{\rm{K}}_{\rm{a}}} + \log \frac{{\left[ {{{\rm{A}}^ - }} \right]}}{{\left[ {{\rm{HA}}} \right]}}$

When volume is same then number of moles is used instead of concentrations of acid and conjugate base. Thus,

${\rm{pH}} = {\rm{p}}{{\rm{K}}_{\rm{a}}} + \log \frac{{{n_{{{\rm{A}}^ - }}}}}{{{n_{{\rm{HA}}}}}}$

Substitute 4.148 for ${\rm{p}}{{\rm{K}}_{\rm{a}}}$, 0.065 mol for ${n_{{{\rm{A}}^ - }}}$and 0.046 mol for ${n_{{\rm{HA}}}}$.

$\begin{array}{c}\\{\rm{pH}} = 4.148 + \log \frac{{0.065}}{{0.046}}\\\\ = 4.30\\\end{array}$

The pH of the resulting buffer is 4.30.

Ans:

The pH of the resulting buffer is 4.30.

Answer 2

The answer is: pH = 4.30. See solution below:

Initial moles of HA = volume x concentration of HA

= 555/1000 x 0.200 = 0.111 mol

Moles of NaOH added = volume x concentration of NaOH

= 500/1000 x 0.130 = 0.065 mol

NaOH + HA => Na+ + A- + H2O

Moles of HA left = initial moles of HA - moles of NaOH added

= 0.111 - 0.065 = 0.046 mol

Moles of A- formed = moles of NaOH added = 0.065 mol

Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

= -log Ka + log(moles of A-/moles of HA) since final volume is the same for both

= -log(7.11 × 10^-5) + log(0.065/0.046)

= 4.30