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Math  /  Algebra

QuestionWhat is the concentration of H+\mathrm{H}^{+}in a 0.54 mol/L0.54 \mathrm{~mol} / \mathrm{L} oxalic acid solution?
Oxalic acid is weak. Therefore, this is an equilibrium problem! HOOCCOOH(aq)+H2O(I)∏H3O+(aq)+HOOCCOO−(aq)Ka=HOOCCOOHH3O+ HOOCCOO I( mol/L)0.5400C( mol/L)−x+x+xE( mol/L)0.54−xxx5.6×10−2=\begin{array}{l} \mathrm{HOOCCOOH}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{I}) \prod \mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})+\mathrm{HOOCCOO}^{-}(\mathrm{aq}) \\ K a= \\ \begin{array}{lllll} \hline & \mathrm{HOOCCOOH} & \mathrm{H}_{3} \mathrm{O}+ & \text { HOOCCOO } \\ \hline \mathrm{I}(\mathrm{~mol} / \mathrm{L}) & 0.54 & 0 & 0 \\ \mathrm{C}(\mathrm{~mol} / \mathrm{L}) & -\mathrm{x} & +\mathrm{x} & +\mathrm{x} \\ \mathrm{E}(\mathrm{~mol} / \mathrm{L}) & 0.54-\mathrm{x} & \mathrm{x} & \mathrm{x} \\ & & 5.6 \times 10^{-2}= & \end{array} \end{array}

Studdy Solution

STEP 1

1. Oxalic acid is a weak acid and partially dissociates in water.
2. The equilibrium expression for the dissociation of oxalic acid is given.
3. The initial concentration of oxalic acid is 0.54 mol/L0.54 \, \text{mol/L}.
4. The equilibrium constant KaK_a for the dissociation is 5.6×10−25.6 \times 10^{-2}.
5. We need to find the concentration of H+\mathrm{H}^{+} ions at equilibrium.

STEP 2

1. Write the equilibrium expression for the dissociation of oxalic acid.
2. Set up the ICE table based on the initial conditions and changes at equilibrium.
3. Substitute the equilibrium concentrations into the equilibrium expression.
4. Solve for xx, which represents the concentration of H+\mathrm{H}^{+} ions.

STEP 3

Write the equilibrium expression for the dissociation of oxalic acid.
The dissociation reaction is: HOOCCOOH(aq)+H2O(l)⇌H3O+(aq)+HOOCCOO−(aq)\mathrm{HOOCCOOH}(\mathrm{aq}) + \mathrm{H}_2\mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_3\mathrm{O}^{+}(\mathrm{aq}) + \mathrm{HOOCCOO}^{-}(\mathrm{aq})
The equilibrium expression is: Ka=[H3O+][HOOCCOO−][HOOCCOOH]K_a = \frac{[\mathrm{H}_3\mathrm{O}^{+}][\mathrm{HOOCCOO}^{-}]}{[\mathrm{HOOCCOOH}]}

STEP 4

Set up the ICE table based on the initial conditions and changes at equilibrium.
Initial concentrations: - [HOOCCOOH]=0.54 mol/L[\mathrm{HOOCCOOH}] = 0.54 \, \text{mol/L} - [H3O+]=0[\mathrm{H}_3\mathrm{O}^{+}] = 0 - [HOOCCOO−]=0[\mathrm{HOOCCOO}^{-}] = 0
Change in concentrations: - [HOOCCOOH]=−x[\mathrm{HOOCCOOH}] = -x - [H3O+]=+x[\mathrm{H}_3\mathrm{O}^{+}] = +x - [HOOCCOO−]=+x[\mathrm{HOOCCOO}^{-}] = +x
Equilibrium concentrations: - [HOOCCOOH]=0.54−x[\mathrm{HOOCCOOH}] = 0.54 - x - [H3O+]=x[\mathrm{H}_3\mathrm{O}^{+}] = x - [HOOCCOO−]=x[\mathrm{HOOCCOO}^{-}] = x

STEP 5

Substitute the equilibrium concentrations into the equilibrium expression.
Ka=(x)(x)0.54−x=5.6×10−2K_a = \frac{(x)(x)}{0.54 - x} = 5.6 \times 10^{-2}
Simplify to get: x2=5.6×10−2×(0.54−x)x^2 = 5.6 \times 10^{-2} \times (0.54 - x)

STEP 6

Solve for xx, which represents the concentration of H+\mathrm{H}^{+} ions.
Assume xx is small compared to 0.54, so 0.54−x≈0.540.54 - x \approx 0.54.
x2=5.6×10−2×0.54x^2 = 5.6 \times 10^{-2} \times 0.54
Calculate: x2=3.024×10−2x^2 = 3.024 \times 10^{-2}
x=3.024×10−2x = \sqrt{3.024 \times 10^{-2}}
x≈0.174x \approx 0.174
The concentration of H+\mathrm{H}^{+} ions is approximately 0.174 mol/L0.174 \, \text{mol/L}.

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