Question eae43

Nov 23, 2017

The percent ionization of salicylic acid is approximately 23.7% according to the given data.

Explanation:

$\frac{2 g}{L} \cdot \frac{m o l}{138 g} \approx 1.45 \cdot {10}^{-} 2 M$

${C}_{7} {H}_{6} {O}_{3} \left(a q\right) r i g h t \le f t h a r p \infty n s {H}^{+} \left(a q\right) + {C}_{7} {H}_{5} {O}_{3}^{-} \left(a q\right)$

${K}_{a} = \frac{\left[{H}^{+}\right] \left[{C}_{7} {H}_{5} {O}_{3}^{-}\right]}{\left[{C}_{7} {H}_{6} {O}_{3}\right]}$

$1.06 \cdot {10}^{-} 3 = {x}^{2} / \left(0.0145 - x\right)$
$\therefore x = \left[{H}^{+}\right] \approx 0.00343 M$

%I = ([H^+])/([C_7H_6O_3]) = (0.00343)/(0.0145M)*100% approx 23.7%

Nov 23, 2017

This is basically an equilibrium problem where you know the initial concentration (a usual one). But your cited $\text{pH}$ is not accurate. I used your ${K}_{a}$ to get a percent dissociation of ul(17.7%).

[Your cited $\text{pH}$ suggests a percent dissociation that is far too low.]

Write the reaction first, so you have a way to organize:

${\text{HC"_7"H"_5"O"_3(aq) rightleftharpoons "C"_7"H"_5"O"_3^(-)(aq) + "H}}^{+} \left(a q\right)$

$\text{I"" "["HC"_7"H"_5"O"_3]_i" "" "" ""0 M"" "" "" "" ""0 M}$
$\text{C"" "" "-x" "" "" "" "" "+x" "" "" } + x$
$\text{E"" "["HC"_7"H"_5"O"_3]_(i) - x" "x" M"" "" "" "" "x" M}$

METHOD 1: USING THE $\boldsymbol{{\text{K}}_{a}}$ (this works)

Next, apparently we know ${\left[{\text{HC"_7"H"_5"O}}_{3}\right]}_{i}$ already (not equilibrium but initial). However, in water at ${25}^{\circ} \text{C}$, the cited solubility is $\text{2.48 g/L}$, so that's what I'll use.

If you want to use a solubility, you must cite its temperature.

$\textcolor{g r e e n}{\left[{\text{HC"_7"H"_5"O"_3]_(i) = (2.48 cancel"g")/"L" xx "1 mol"/(138.12 cancel("g H"_7"H"_5"O}}_{3}\right)}$

$=$ $\textcolor{g r e e n}{\text{0.0179"_6 "M}}$

Using your cited ${K}_{a}$, which is actually correct, we can do this in one of two ways. We know ${\left[{\text{HC"_7"H"_5"O}}_{3}\right]}_{i}$, so...

${K}_{a} = 1.06 \times {10}^{- 3} = {x}^{2} / \left(\text{0.01796 M} - x\right)$

${x}^{2} + 1.06 \times {10}^{- 3} x - 1.06 \times {10}^{- 3} \cdot 0.01796 = 0$,

and the result is x = ul(["H"^(+)]_(eq) = "0.00387 M") from the quadratic formula.

The percent dissociation is

color(blue)(%"dissoc") = x/(["HA"]_i) xx 100%

= ("0.00387 M")/("0.0179"_6 "M" + "0.00387 M") xx 100%

= color(blue)ul(17.7%)

And that's reasonable. Salicylic acid has a non-negligible percent dissociation with a ${K}_{a}$ larger than ${10}^{- 5}$.

METHOD 2: TRYING THE pH (this won't work)

With the $\text{pH}$,

${10}^{- \text{pH") = ul(["H"^(+)]_(eq)) = x = 10^(-4.96) "M" = ul(1.10 xx 10^(-5) "M}}$

As before, the solubility was $\text{0.01796 M}$.

This is pretty big in comparison to $x$. As a result, the percent dissociation using the cited $\text{pH}$ will be far too low.

color(red)(%"dissoc") = x/(["HA"]_i) xx 100%

= (10^(-4.96) "M")/("0.0179"_6 "M") xx 100% = color(red)(0.0611%)#

which doesn't make sense how low this is, given the size of ${K}_{a}$.

In fact, the ${K}_{a}$ obtained from this percent dissociation is

$\textcolor{red}{{K}_{a}} = \left({10}^{- 4.96} \text{M")^2/("0.0179"_6 "M" - 10^(-4.96)"M}\right) = \textcolor{red}{6.70 \times {10}^{- 9}}$,

which is far from the actual.