Question

A buffer solution of pH = 5.15 is to be prepared from acetic acid (Ka =

$1.80x10^5$ ), sodium acetate and water. What must be the molar ratio of acetate ion to acetic acid in this solution to obtain this pH?

• A. $0.05mole$
• B. $2.54mole$
• C. $0.1mole$
• D. $0.005mole$

Answer 1

The correct option is B $2.54mole$

$\begin{array}{c}\text{To create:-buffer of}pH=5.15\text{from}\\ \text{acctic acid}(ka=1.8\times {10}^{-5})\text{, sodium acetate}\\ \text{and water.}\\ \text{To find:- molar ratio of acetate ion to}\\ \text{acetic acid}\\ \text{According to Henderson- Hassel batch equation}\\ pH=pka+\log \left[\frac{\text{salt}}{\text{acid}}\right].\end{array}$

$\text{Here}pka=-\log (1\cdot 8\times {10}^{-5})=4\cdot 7447$

$\begin{array}{cc}\therefore & pH-pka={\log }_{10}\left[\frac{\left[acetate\right]}{\left[acid\right]}\right]\\ & 5\cdot 15-4\cdot 7447={\log }_{10}[\frac{\left(a\text{cetate ion}\right)}{\text{(acetic acid)}}\end{array}$

${10}^{0.3753}=\frac{\text{[acetate ion]}}{\text{[acetic acid]}}$

[acetate] $\frac{\text{ion}]}{\text{[acetic acid]}}=2\cdot 54$

So, the correct answer is B) 2.54