# Question #f7626

Apr 14, 2016

$\left(a\right) \text{ "1.44"g}$

$\left(b\right) \text{ ""pH"" = } 4.1$

#### Explanation:

$\left(a\right)$

I will refer to propanoic acid as $H A$. It is a weak acid and dissociates as follows:

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

For which:

${K}_{a} = \frac{\left[{H}_{\left(a q\right)}^{+}\right] \left[{A}_{\left(a q\right)}^{-}\right]}{\left[H {A}_{\left(a q\right)}\right]} = 1.35 \times {10}^{- 5} \text{mol/l}$

The concentrations are at equilibrium. Rearranging gives:

$\left[{H}_{\left(a q\right)}^{+}\right] = {K}_{a} \times \frac{\left[H {A}_{\left(a q\right)}\right]}{\left[{A}_{\left(a q\right)}^{-}\right]} \text{ } \textcolor{red}{\left(1\right)}$

We are told that $p H = 7.4 = - \log \left[{H}_{\left(a q\right)}^{+}\right]$

From which $\left[{H}_{\left(a q\right)}^{+}\right] = 1.78 \times {10}^{- 5} \text{mol/l}$

$\therefore 1.78 \times {10}^{- 5} = 1.34 \times {10}^{- 5} \times \frac{\left[H {A}_{\left(a q\right)}\right]}{\left[{A}_{\left(a q\right)}^{-}\right]}$

$\therefore \frac{\left[H {A}_{\left(a q\right)}\right]}{\left[{A}_{\left(a q\right)}^{-}\right]} = \frac{1.78 \times \cancel{{10}^{- 5}}}{1.34 \times \cancel{{10}^{- 5}}} = 1.33$

The total volume is common to both $H A$ and ${A}^{-}$ so we can write moles $n$ instead of concentration.

$\therefore \frac{n H A}{n {A}^{-}} = 1.33$

I will assume that we have $0.02 \text{mol }$ of $H A$ at equilibrium as the amount that dissociates is very small in comparison.

$\therefore n {A}^{-} = \frac{0.02}{1.33} = 0.015$

The ${M}_{r}$ of sodium propanoate $= 96.06$

So the mass needed is given by:

$m = 0.015 \times 96.06 = 1.44 \text{g}$

Again, I have neglected the tiny amount of ${A}^{-}$ formed from the dissociation of the acid.

There may be a volume change on mixing but again, this is common to acid and co-base so will cancel.

$\left(b\right)$

The buffer works because there is a large reserve of the co-base ${A}^{-}$ which can absorb the addition of small amounts of ${H}^{+}$ ions:

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

The position of equilibrium lies well to the right.

Since they react in a 1:1 molar ratio we can see that adding $0.01 \text{mol}$ of ${H}^{+}$ will consume $0.01 \text{mol}$ of ${A}^{-}$ and form $0.01 \text{mol}$ of $H A$.

$\therefore n {A}^{-}$ remaining $= 0.015 - 0.01 = 0.005$

The no. moles $H A$ will increase by $0.01$ so:

$n H A = 0.02 + 0.01 = 0.03$

Using the usual assumptions already described we can insert these values into $\textcolor{red}{\left(1\right)} \Rightarrow$

$\left[{H}_{\left(a q\right)}^{+}\right] = 1.35 \times {10}^{- 5} \times \frac{0.03}{0.005}$

$\left[{H}_{\left(a q\right)}^{+}\right] = 8.1 \times {10}^{- 5} \text{mol/l}$

$\therefore p H = - \log \left[8.1 \times {10}^{- 5}\right]$

$\textcolor{red}{\text{pH=4.1}}$

We are asked to compare this with the pH of $0.01 \text{M"" " "HCl}$ solution.

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

The $\text{HCl}$ solution can be assumed to be 100% dissociated so:

$p H = - \log \left(0.01\right)$

$\textcolor{red}{\text{pH=2}}$

This is, therefore, considerably more acidic when not made up in a buffer.