# Question #9da57

Mar 27, 2016

You would need $0.00435$ moles of $\text{HCl}$.

#### Explanation:

Hydrocyanic acid is a weak acid that ionises:

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

For which:

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

These are equilibrium concentrations.

Rearranging gives:

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

We can use the data given to get the acid/salt ratio.

$p H = 8.5$

$\therefore \log \left[{H}_{\left(a q\right)}^{+}\right] = - 8.5$

From which $\left[{H}_{\left(a q\right)}^{+}\right] = 3.16 \times {10}^{- 9} \text{mol/l}$

Putting in the numbers $\Rightarrow$

$3.16 \times {10}^{- 9} = 4.1 \times {10}^{- 10} \times \frac{\left[H C {N}_{\left(a q\right)}\right]}{\left[C {N}_{\left(a q\right)}^{-}\right]}$

$\therefore \frac{\left[H C {N}_{\left(a q\right)}\right]}{\left[C {N}_{\left(a q\right)}^{-}\right]} = \frac{3.16 \times {10}^{- 9}}{4.1 \times {10}^{- 10}} = 0.77$

Because ${K}_{a}$ is so small we can say that the equilibrium shown above lies well to the left.

We can therefore assume that if ${H}^{+}$ ions are added to a solution containing $C {N}^{-}$ (nitrile) ions then they will form $H C N$ and we can ignore the tiny amount that dissociate.

So if $n$ moles of ${H}^{+}$ ions are added to a solution containing $0.01$ moles of $C {N}^{-}$ then $n$ moles of $H C N$ form and $0.01 - n$ moles of $C {N}^{-}$ remain.

Since the total volume is common to both $H C N$ and $C {N}^{-}$ we can write:

$\frac{n}{0.01 - n} = 0.77$

$\therefore n = 0.77 \left(0.01 - n\right)$

$\therefore n = 0.0077 - 0.77 n$

$\therefore n = \frac{0.0077}{1.77} = 0.00435 \text{mol}$

This is the number of moles of $H C l$ needed.

The question does not specify any concentration of $H C l$ to be used so here's a possible recipe for the buffer:

Say we are given $H C l$ of concentration $1 \text{mol/l}$

How much would we need?

$c = \frac{n}{v} \therefore v = \frac{n}{c} = \frac{0.00435}{1} = 0.00435 \text{L"=4.35"ml}$

The ${M}_{r}$ of $N a C N = 49.01$

$\therefore 0.01 \text{mol"=0.49"g}$

You could dissolve this in a minimum amount of distilled water and transfer, with washings, to a 1 litre volumetric flask.

Then add $4.35 \text{ml}$ of $1 \text{M""" } H C l$ from a graduated pipette and make up to the 1 litre mark.