Question #b41c8

Question

Question #b41c8

1 Answer

Impact of this question

1733 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-06T00:36:01

Here's what I got.

Explanation:

You know that a weak monoprotic acid will only partially ionize to produce hydronium cations and the conjugate base of the acid in $1 : 1$ $1:1$ mole ratios, so if you take $HA$ $"HA"$ to be your generic monoprotic weak acid, you can say that you have

${HA}_{(a q)} + H_{2} O_{(l)} ⇌ H_{3} O_{(a q)}^{+} + A_{(a q)}^{-}$ $"HA"_ ((aq)) + "H"_ 2"O"_ ((l)) rightleftharpoons "H"_ 3"O"_ ((aq))^(+) + "A"_ ((aq))^(-)$

Now, you know that you start with $1.75$ $1.75$ moles of weak acid and that, at equilibrium, the solution contains $1.15$ $1.15$ moles of weak acid.

This implies that

$175 moles - 1.15 moles = 0.60 moles$ $"175 moles " - " 1.15 moles" = "0.60 moles"$

of weak acid have ionized to produce hydronium cations and the conjugate base of the weak acid.

This implies that, at equilibrium, the solution contains $0.60$ $0.60$ moles of hydronium cations and $0.60$ $0.60$ moles of $A^{-}$ $"A"^(-)$ $\to$ $->$ this is the case because for every $1$ $1$ mole of weak acid that dissociates, you get $1$ $1$ mole of each product.

Use the volume of the solution to calculate the equilibrium concentrations of the species involved in the reaction.

$[HA] = \frac{1.15 moles}{1.5 L} = {0.767 mol L}^{- 1}$ $["HA"] = "1.15 moles"/"1.5 L" = "0.767 mol L"^(-1)$

$[H_{3} O^{+}] = \frac{0.60 moles}{1.5 L} = {0.40 moles L}^{- 1}$ $["H"_3"O"^(+)] = "0.60 moles"/"1.5 L" = "0.40 moles L"^(-1)$

$[A^{-}] = \frac{0.60 moles}{1.5 L} = {0.40 moles L}^{- 1}$ $["A"^(-)] = "0.60 moles"/"1.5 L" = "0.40 moles L"^(-1)$

By definition, the acid dissociation constant, $K_{a}$ $K_a$ , which uses equilibrium concentrations, is equal to

$K_{a} = \frac{[H_{3} O^{+}] \cdot [A^{-}]}{[HA]}$ $K_a = (["H"_3"O"^(+)] * ["A"^(-)])/(["HA"])$

Plug in your values to find--I'll leave the answer without added units

$K_a = (0.40 * 0.40)/(0.767) = color(darkgreen)(ul(color(black)(0.21)))$

The answer is rounded to two sig figs, the number of sig figs you have for the volume of the solution.

Finally, the $"pH"$ of the solution is equal to

$color(blue)(ul(color(black)("pH" = - log(["H"_3"O"^(+)]))))$

In your case, the solution will have a $"pH"$ of

$"pH" = - log(0.40) = color(darkgreen)(ul(color(black)(0.40)))$

The answer is rounded to two decimal places because you have two sig figs for the volume of the solution.

Now, notice that the $"pH"$ of the solution is quite low. However, you can say that you're dealing with a weak acid because strong acids are characterized by the fact that they ionize completely, i.e. $1$ mole of acid produces $1$ mole of hydronium cations, in aqueous solution to produce hydronium cations.

In your case, the acid does not ionize completely, i.e. $1$ mole of acid does not produce $1$ mole of hydronium cations, so you can classify it as a weak acid despite the very low $"pH"$ value of its solution.

Science

Math

Humanities

... and beyond

Question #b41c8

1 Answer

Explanation:

Related questions

Impact of this question