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Chapter 14: Problem 74

A solution contains a mixture of acids: 0.50\(M\) HA \(\left(K_{\mathrm{a}}=1.0\right.\) $\times 10^{-3} ), 0.20 M \mathrm{HB}\left(K_{\mathrm{a}}=1.0 \times 10^{-10}\right),\( and 0.10 \)\mathrm{MHC}\left(K_{\mathrm{a}}=\right.\( \)1.0 \times 10^{-12} ) .\( Calculate the \)\left[\mathrm{H}^{+}\right]$ in this solution.

Short Answer

Expert verified
The total concentration of H+ ions in the solution is \([H+]_{total} = [H+]_{HA} + [H+]_{HB} + [H+]_{HC} = \sqrt{(1.0 \times 10^{-3})(0.50)} + \sqrt{(1.0 \times 10^{-10})(0.20)} + \sqrt{(1.0 \times 10^{-12})(0.10)}\). Calculate the individual values and add them up to find the total [H+].

Step by step solution

01

Determine dissociation of HA

First, let's determine how much HA will dissociate in the solution. The Ka of HA is given as \(1.0 \times 10^{-3}\). To find the H+ concentration contributed by HA, we will use the expression for Ka: Ka = \(\frac{[H+][A-]}{[HA]}\) Where: - [H+] is the concentration of hydrogen ions - [A-] is the concentration of the conjugate base - [HA] is the concentration of the acid Since HA is a weak acid, its dissociation will be relatively small, so we can make the assumption: [H+] = [A-] and [HA] ≈ 0.50M which is the initial concentration Ka = \(\frac{[H+]^2}{0.50}\) Solving for [H+], we get: [H+] = \(\sqrt{Ka \times [HA]}\) [H+] = \(\sqrt{(1.0 \times 10^{-3})(0.50)}\) #Step 2: Calculate [H+] from HB#
02

Determine dissociation of HB

Next, we will determine how much HB will dissociate in the solution. The Ka of HB is given as \(1.0 \times 10^{-10}\). Using the same process as in Step 1: Ka = \(\frac{[H+][B-]}{[HB]}\) Making the assumption that [H+] = [B-] and [HB] ≈ 0.20M : Ka = \(\frac{[H+]^2}{0.20}\) Solving for [H+]: [H+] = \(\sqrt{Ka \times [HB]}\) [H+] = \(\sqrt{(1.0 \times 10^{-10})(0.20)}\) #Step 3: Calculate [H+] from HC#
03

Determine dissociation of HC

Lastly, we will determine how much HC will dissociate in the solution. The Ka of HC is given as \(1.0 \times 10^{-12}\). Using the same process as in previous steps: Ka = \(\frac{[H+][C-]}{[HC]}\) Making the assumption that [H+] = [C-] and [HC] ≈ 0.10M : Ka = \(\frac{[H+]^2}{0.10}\) Solving for [H+]: [H+] = \(\sqrt{Ka \times [HC]}\) [H+] = \(\sqrt{(1.0 \times 10^{-12})(0.10)}\) #Step 4: Calculate the total [H+]#
04

Adding the H+ contributions from HA, HB, and HC

Now that we have calculated the H+ concentration contributed by each of the acids HA, HB, and HC, we can simply add them up to determine the total H+ concentration in the solution: \([H+]_{total} = [H+]_{HA} + [H+]_{HB} + [H+]_{HC}\) Calculate the values obtained from the previous steps and add them up to find the total concentration of H+ ions in the solution.

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Most popular questions from this chapter

A \(0.100-\mathrm{g}\) sample of the weak acid \(\mathrm{HA}\) (molar mass \(=\) 100.0 \(\mathrm{g} / \mathrm{mol} )\) is dissolved in 500.0 \(\mathrm{g}\) water. The freezing point of the resulting solution is \(-0.0056^{\circ} \mathrm{C}\) . Calculate the value of \(K_{\mathrm{a}}\) for this acid. Assume molality equals molarity in this solution.

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=378.85 \mathrm{g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=\) \(8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C} ) .\) Calculate the \(\mathrm{pH}\) of a \(30.0-\mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at $35.0^{\circ} \mathrm{C}$ is \(2.1 \times 10^{-14} .\)

Trichloroacetic acid $\left(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}\right)$ is a corrosive acid that is used to precipitate proteins. The pH of a \(0.050-M\) solution of trichloroacetic acid is the same as the pH of a \(0.040-M \mathrm{HClO}_{4}\) solution. Calculate \(K_{\mathrm{a}}\) for trichloroacetic acid.

A \(1.0 \times 10^{-2}-M\) solution of cyanic acid (HOCN) is 17\(\%\) dissociated. Calculate \(K_{\mathrm{a}}\) for cyanic acid.

Isocyanic acid \((\mathrm{HNCO})\) can be prepared by heating sodium cyanate in the presence of solid oxalic acid according to the equation $$ 2 \mathrm{NaOCN}(s)+\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) \longrightarrow 2 \mathrm{HNCO}(l)+\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}(s) $$ Upon isolating pure HNCO \((l),\) an aqueous solution of HNCO can be prepared by dissolving the liquid HNCO in water. What is the pH of a 100 -mL solution of HNCO prepared from the reaction of 10.0 g each of NaOCN and $\mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4},$ assuming all of the HNCO produced is dissolved in solution? \(\left(K_{\mathrm{a}} \text { of HNCO }\right.\) \(=1.2 \times 10^{-4} . )\)
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