Question c1d7c

Mar 30, 2016

$\left(a\right) . p H = 4.56$

$\left(b\right)$. An extra $4.85 \text{ml}$ must be added.

Explanation:

$\left(a\right) .$

Ethanoic acid is a weak acid and dissociates:

$C {H}_{3} C O O {H}_{\left(a q\right)} r i g h t \le f t h a r p \infty n s C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{\left(a q\right)}^{+} \text{ } \textcolor{red}{\left(1\right)}$

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

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

These refer to equilibrium concentrations.

When the $N a O {H}_{\left(a q\right)}$ is added the following neutralisation takes place:

$C {H}_{3} C O O {H}_{\left(a q\right)} + O {H}_{\left(a q\right)}^{-} \rightarrow C {H}_{3} C O {O}_{\left(a q\right)}^{-} + {H}_{2} {O}_{\left(l\right)} \text{ } \textcolor{red}{\left(3\right)}$

This tells us that for every mole of $O {H}_{\left(a q\right)}^{-}$ added, 1 mole of $C {H}_{3} C O {O}_{\left(a q\right)}^{-}$ will form and 1 mole of $C {H}_{3} C O O {H}_{\left(a q\right)}$ will be consumed.

The number of moles of $O {H}_{\left(a q\right)}^{-}$ added is given by:

${n}_{O {H}^{-}} = c \times v = 0.2 \times \frac{20}{1000} = 4 \times {10}^{- 3}$

From $\textcolor{red}{\left(3\right)}$ we can therefore say:

${n}_{C {H}_{3} C O {O}^{-}} = 4 \times {10}^{-} 3$

The initial moles of $C {H}_{3} C O O {H}_{\left(a q\right)}$ is given by:

Initial moles $= c \times v = 0.2 \times \frac{50}{1000} = 10 \times {10}^{- 3}$

So the number of moles of $C {H}_{3} C O O {H}_{\left(a q\right)}$ remaining is given by:

${n}_{C {H}_{3} C O O H} = \left(10 \times {10}^{- 3}\right) - \left(4 \times {10}^{- 3}\right) = 6 \times {10}^{- 3}$

Now we can set up an ICE table:

$C {H}_{3} C O O {H}_{\left(a q\right)} \text{ "rightleftharpoons" "CH_3COO_((aq))^(-)" } + {H}_{\left(a q\right)}^{+}$

color(red)("I")" "6xx10^(-3)" "4xx10^(-3)" "0

color(red)("C")" "-x" "+x" "+x

color(red)("E")" "(6xx10^(-3)-x)" "(4xx10^(-3)+x)" "x

Because the dissociation is so small I am going to assume that $x$ is much smaller than the initial moles of both $C {H}_{3} C O O H$ and $C {H}_{3} C O {O}^{-}$. This means that:

$\left(6 \times {10}^{- 3} - x\right) \Rightarrow 6 \times {10}^{- 3}$

and

$\left(4 \times {10}^{- 3} - x\right) \Rightarrow 4 \times {10}^{- 3}$

So rearranging $\textcolor{red}{\left(2\right)}$ gives:

$\left[{H}^{+}\right] = {k}_{a} \times \frac{\left[C {H}_{3} C O O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]} \text{ } \textcolor{red}{\left(4\right)}$

Since the total volume is common we can write:

$\left[{H}^{+}\right] = 1.8 \times {10}^{- 5} \times \left[\frac{6 \times \cancel{{10}^{- 3}}}{4 \times \cancel{{10}^{- 3}}}\right]$

$\therefore \left[{H}^{+}\right] = 1.8 \times {10}^{- 5} \times \frac{3}{2} = 2.7 \times {10}^{- 5}$

$p H = - \log \left(2.7 \times {10}^{- 5}\right)$

color(red)(pH=4.56#

$\left(b\right) .$

Now the pH is raised by adding extra alkali. We can use the new pH to get the ratio of acid to salt:

$p H = 4.74$

$\therefore - \log \left[{H}^{+}\right] = 4.74$

$\therefore \left[{H}^{+}\right] = 1.82 \times {10}^{- 5} \text{mol/l}$

From $\textcolor{red}{\left(4\right)}$ we can write:

$1.82 \times \cancel{{10}^{- 5}} = 1.8 \times \cancel{{10}^{- 5}} \times \frac{\left[C {H}_{3} C O O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]}$

$\therefore \frac{\left[C {H}_{3} C O O H\right]}{\left[C {H}_{3} C O {O}^{-}\right]} = \frac{1.82}{1.8} = 1.01 \text{ } \textcolor{red}{\left(5\right)}$

The number of moles of $O {H}^{-}$ is given by:

${n}_{O {H}^{-}} = 0.2 \times {V}_{O {H}^{-}}$

$\therefore {n}_{C {H}_{3} C O {O}^{-}} = 0.2 \times {V}_{O {H}^{-}}$

and

${n}_{C {H}_{3} C O O H} = \left(10 \times {10}^{- 3}\right) - 0.2 \times {V}_{O {H}^{-}}$

Substituting these into $\textcolor{red}{\left(5\right)}$ gives:

$\frac{0.01 - 0.2 {V}_{O {H}^{-}}}{0.2 {V}_{O {H}^{-}}} = 1.01$

$\therefore 1.01 \times 0.2 {V}_{O {H}^{-}} = 0.01 - 0.2 {V}_{O {H}^{-}}$

$\therefore 0.202 {V}_{O {H}^{-}} + 0.2 {V}_{O {H}^{-}} = 0.01$

$\therefore 0.402 {V}_{O {H}^{-}} = 0.01$

$\therefore {V}_{O {H}^{-}} = \frac{0.01}{0.402} = 0.02487 \text{L}$

${V}_{O {H}^{-}} = 24.87 \text{ml}$

Since $20 \text{ml}$ was used initially, this represents an extra $4.87 \text{ml}$ of $0.2 \text{M"" } N a O {H}_{\left(a q\right)}$

In practice this would be $4.85 \text{ml}$ given the usual accuracy of a graduated pipette.