# A buffer solution of pH = 5.15 is to be prepared from acetic acid (Ka = 1.80 x 10^-5), sodium acetate and water. What must be the molar ratio of acetate ion to acetic acid in this solution to obtain this pH?

Jun 27, 2016

$2.54$

#### Explanation:

Ethanoic acid is a weak acid which dissociates:

$C {H}_{3} C O O {H}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{\left(a q\right)}^{+}$

The expression for ${K}_{a}$ is:

${K}_{a} = \frac{\left[C {H}_{3} C O {O}_{\left(a q\right)}^{-}\right] \left[{H}_{\left(a q\right)}^{+}\right]}{\left[C {H}_{3} C O O {H}_{\left(a q\right)}\right]}$

So the molar ratio of acetate ion to acetic acid is:

$\frac{\left[C {H}_{3} C O {O}_{\left(a q\right)}^{-}\right]}{\left[C {H}_{3} C O O {H}_{\left(a q\right)}\right]} = {K}_{a} / \left[\left[{H}_{\left(a q\right)}^{+}\right]\right]$

The ratio of conentrations which I have given also corresponds to the molar ratio since the volume is common to both.

We are told that the $p H = 5.15$

$\therefore - \log \left[{H}_{\left(a q\right)}^{+}\right] = 5.15$

$\therefore \left[{H}_{\left(a q\right)}^{+}\right] = 7.08 \times {10}^{- 6} \text{ ""mol/l}$

$\therefore \frac{\left[C {H}_{3} C O {O}_{\left(a q\right)}^{-}\right]}{\left[C {H}_{3} C O O {H}_{\left(a q\right)}\right]} = \frac{n C {H}_{3} C O {O}^{-}}{n C {H}_{3} C O O H} = \frac{1.8 \times {10}^{- 5}}{7.08 \times {10}^{- 6}} = 2.54$