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Chapter 14: Problem 75

The \(\mathrm{pH}\) of a \(1.00 \times 10^{-2} \mathrm{M}\) solution of cyanic acid (HOCN) is \(2.77\) at \(25^{\circ} \mathrm{C}\). Calculate \(K_{\mathrm{a}}\) for \(\mathrm{HOCN}\) from this result.

Short Answer

Expert verified
Given the pH of $2.77$, we can calculate the [H+]: \([H+] = 10^{-2.77} = 1.68 \times 10^{-3} \mathrm{M}\). We find the equilibrium concentrations: [HOCN] = \(8.32 \times 10^{-3} \mathrm{M}\), [H+] = \(1.68 \times 10^{-3} \mathrm{M}\), and [OCN-] = \(1.68 \times 10^{-3} \mathrm{M}\). Using the equilibrium expression: \(K_a = [H+][OCN^-] / [HOCN] = 3.37 \times 10^{-4}\).

Step by step solution

01

Calculate the concentration of H+ ions

Given the pH of the solution, we can determine the concentration of H+ ions using the formula: pH = -log[H+] Rewrite the formula for [H+]: [H+] = 10^(-pH) Now we will plug in the given pH value: [H+] = 10^(-2.77) = \(1.68 \times 10^{-3} \mathrm{M}\) Step 2: Find equilibrium concentrations of all species
02

Calculate equilibrium concentrations of HOCN, H+, and OCN-

Since the initial concentration of HOCN is 1.00 × 10^-2 M and the change in the concentration of the H+ ions is equal to that of HOCN, we can deduce the equilibrium concentrations. [HOCN] = initial concentration - change in concentration = 1.00 × 10^-2 M - 1.68 × 10^-3 M = \(8.32 \times 10^{-3} \mathrm{M}\) [H+] = 1.68 × 10^-3 M (as calculated earlier) [OCN-] = 1.68 × 10^-3 M (since it is proportional to H+) Step 3: Calculate Ka
03

Use the equilibrium expression to calculate Ka

Now that we have the equilibrium concentrations of all species, we can use the equilibrium expression to find Ka: Ka = [H+][OCN-] / [HOCN] Ka = (1.68 × 10^-3)(1.68 × 10^-3) / (8.32 × 10^(-3)) Ka = \(3.37 \times 10^{-4}\) Thus, the Ka for HOCN is \(3.37 \times 10^{-4}\).

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Most popular questions from this chapter

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 \(2.1 \times 10^{-14}\).

Arsenic acid \(\left(\mathrm{H}_{3} \mathrm{AsO}_{4}\right)\) is a triprotic acid with \(K_{\mathrm{a}_{1}}=5 \times 10^{-3}\) \(K_{\mathrm{a}_{2}}=8 \times 10^{-8}\), and \(K_{\mathrm{a}_{3}}=6 \times 10^{-10} \cdot\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\) \(\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right],\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right],\left[\mathrm{HAsO}_{4}^{2-}\right]\), and \(\left[\mathrm{AsO}_{4}^{3-}\right]\) in a \(0.20 \mathrm{M}\) arsenic acid solution.

Calculate the \(\mathrm{pH}\) of a \(0.200 \mathrm{M}\) solution of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\). Hint: \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NHF}\) is a salt composed of \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\) and \(\mathrm{F}^{-}\) ions. The prin- cipal equilibrium in this solution is the best acid reacting with the best base; the reaction for the principal equilibrium is \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}(a q)+\mathrm{F}^{-}(a q) \rightleftharpoons\) \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N}(a q)+\mathrm{HF}(a q) \quad K=8.2 \times 10^{-3}\)

Will \(0.10 M\) solutions of the following salts be acidic, basic, or neutral? See Appendix 5 for \(K_{\mathrm{a}}\) values. a. ammonium bicarbonate b. sodium dihydrogen phosphate c. sodium hydrogen phosphate d. ammonium dihydrogen phosphate e. ammonium formate

What are the major species present in \(0.250 \mathrm{M}\) solutions of each of the following acids? Calculate the \(\mathrm{pH}\) of each of these solutions. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)
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