# What is an example of a pH buffer calculation problem?

Apr 12, 2014

For most buffer calculations, you use the Henderson-Hasselbalch equation

pH = $p {K}_{a} + \log \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$ for acids or

pOH = $p {K}_{b} + \log \left(\frac{\left[B {H}^{+}\right]}{\left[B\right]}\right)$ for bases

Example 1

Calculate the pH of a buffer solution made from 0.20 mol/L HC₂H₃O₂ and 0.50 mol/L C₂H₃O₂⁻. The acid dissociation constant of HC₂H₃O₂ is 1.8 × 10⁻⁵.

Solution

Plug the values into the Henderson-Hasselbalch equation for a weak acid and its conjugate base.

HA + H₂O → H₃O⁺ + A⁻

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

pH = -log (1.8 × 10⁻⁵) + $\log \left(\left(0.50 \text{ mol/L")/(0.20" mol/L}\right)\right)$

pH = -log (1.8 × 10⁻⁵) + log (2.5)

pH = 4.74 + 0.40

pH = 5.14

Example 2

How many moles of sodium acetate and acetic acid must you use to prepare 1.00 L of a 0.100 mol/L buffer with pH 5.00.

Solution

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

5.00 = 4.74 + $\log \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$

$\log \left(\frac{\left[{A}^{-}\right]}{\left[H A\right]}\right)$ = 0.26

$\frac{\left[{A}^{-}\right]}{\left[H A\right]} = {10}^{0.26}$ = 1.82

[A⁻] = 1.82[HA]

Also, [A⁻] + [HA] = 0.100 mol/L

1.82[HA] + [HA] = 0.100 mol/L

2.82[HA] = 0.100 mol/L

[HA] = 0.0355 mol/L

[A⁻] = (0.100 – 0.0355) mol/L = 0.0645 mol/L

You need 0.0355 mol of acetic acid and 0.0645 mol of sodium acetate to prepare 1 L of the buffer.

More than one problem with solutions with additional info here.