# Question #af0cd

Feb 16, 2017

This can be done using Henderson Hasselbalch equation

The equation is

$\textcolor{b l u e}{p H = p {K}_{C {H}_{3} C O O H} + \log \left(\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]}\right)}$

Let us consider that $x m L$ 0.1M sodium Acetate is to be mixed with $\left(200 - x\right) m L$ 2M acetic acid solution to prepare 200ml or 0.2L buffer solution of $p H = 4.50$

So $\left[C {H}_{3} C O O N a\right] \times 0.2 = x \times {10}^{-} 3 \times 0.1$

And $\left[C {H}_{3} C O O H\right] \times 0.2 = \left(200 - x\right) \times {10}^{-} 3 \times 2$

Hence $\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]} = \frac{x}{\left(200 - x\right) \times 20}$

Again $p {K}_{C {H}_{3} C O O H} = 4.75$

Inserting these in Henderson Hasselbalch equation

$\textcolor{b l u e}{p H = p {K}_{C {H}_{3} C O O H} + \log \left(\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]}\right)}$

$\textcolor{b l u e}{\implies 4.50 = 4.75 + \log \left(\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]}\right)}$

$\textcolor{b l u e}{\implies \log \left(\frac{\left[C {H}_{3} C O O N a\right]}{\left[C {H}_{3} C O O H\right]}\right) = 4.5 - 4.75 = - 0.25}$

$\textcolor{b l u e}{\implies \log \left(\frac{x}{\left(200 - x\right) \times 20}\right) = - 0.25}$

$\textcolor{b l u e}{\implies \left(\frac{x}{\left(200 - x\right) \times 20}\right) = {10}^{- 0.25}}$

$\textcolor{b l u e}{\implies \left(\frac{\left(200 - x\right) \times 20}{x}\right) = {10}^{0.25}}$

$\textcolor{b l u e}{\implies \left(\frac{200}{x} - 1\right) = \frac{1}{20} \times {10}^{0.25}}$

$\textcolor{b l u e}{\implies \frac{200}{x} = 1 + \frac{1}{20} \times {10}^{0.25}}$

$\textcolor{red}{\implies x = 183.7 m l}$

So volume of $C {H}_{3} C O O N a = 183.7 m l$

And volume of $C {H}_{3} C O O H = \left(200 - 183.7\right) m L = 16.3 m L$