Question #a2d68

Mar 28, 2016

$40 \text{ml}$ should be added.

Explanation:

The 1st dissociation of carbonic acid is given by:

${H}_{2} C {O}_{3} r i g h t \le f t h a r p \infty n s H C {O}_{3}^{-} + {H}^{+}$

For which:

${K}_{\left(a p p\right)} = \frac{\left[H C {O}_{3}^{-}\right] \left[{H}^{+}\right]}{\left[{H}_{2} C {O}_{3}\right]}$

${K}_{\left(a p p\right)}$ is the value which takes into account the dissolved carbon dioxide. It is equal to $4.0 \times {10}^{- 7} \text{mol/l}$ at ${25}^{\circ} \text{C}$.

The value given in the question looks incorrect. It seems very big.

I will not use the 2nd dissociation.

Rearranging gives:

$\left[{H}^{+}\right] = {K}_{\left(a p p\right)} \times \frac{\left[{H}_{2} C {O}_{3}\right]}{\left[H C {O}_{3}^{-}\right]}$

$\therefore \frac{\left[{H}_{2} C {O}_{3}\right]}{\left[H C {O}_{3}^{-}\right]} = \frac{{H}^{+}}{K} _ \left(\left(a p p\right)\right)$

$p H = 7.4 \therefore \left[{H}^{+}\right] = 4.0 \times {10}^{- 8} \text{mol/l}$

$\therefore \frac{\left[{H}_{2} C {O}_{3}\right]}{\left[H C {O}_{3}^{-}\right]} = \frac{4.0 \times {10}^{- 8}}{4.0 \times {10}^{- 7}} = 0.1$

Since the total volume of the buffer is common to both we can write:

${n}_{{H}_{2} C {O}_{3}} / {n}_{H C {O}_{3}^{-}} = 0.1$

Where $n$ refers to the number of moles.

We know that $c = \frac{n}{v}$ so the number of moles is given by:

$n = c \times v$

So we can write:

$\frac{2 \times \frac{10}{1000}}{{5}_{{V}_{H C {O}_{3}^{-}}}} = 0.1$

$\therefore 5 {V}_{H C {O}_{3}^{-}} = \frac{0.02}{0.1}$

${V}_{H C {O}_{3}^{-}} = \frac{0.02}{5 \times 0.1} = 0.04 \text{L}$

$= 40 \text{ml}$