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Chapter 14: Problem 84

An acid \(\mathrm{HX}\) is 25\(\%\) dissociated in water. If the equilibrium concentration of \(\mathrm{HX}\) is \(0.30 \mathrm{M},\) calculate the \(K_{\mathrm{a}}\) value for \(\mathrm{HX}\) .

Short Answer

Expert verified
The Ka value for the acid HX is 0.025.

Step by step solution

01

Write the dissociation reaction for the acid

For a generic acid, HX, the dissociation reaction in water is : HX (aq) <=> H+ (aq) + X- (aq)
02

Determine the initial, change, and equilibrium concentrations of all species in the reaction using the percent dissociation and ICE table

Given that the acid is 25% dissociated, and the total concentration of HX is 0.3 M, we can determine the concentrations of all species in the reaction: Percent dissociation = (dissociated HX / initial HX) * 100 25% = (dissociated HX / 0.3 M) * 100 We can determine the dissociated HX: dissociated HX = 0.25 * 0.3 M = 0.075 M Since HX dissociates into H+ and X-, each dissociated HX contributes to an equal amount of both ions. Therefore, the initial concentrations of H+ and X- are 0 and 0, respectively. Now we can set up the ICE table: ``` Initial Change Equilibrium HX 0.3 M -0.075 M 0.225 M H+ 0 +0.075 M 0.075 M X- 0 +0.075 M 0.075 M ```
03

Write the Ka expression and solve for Ka

We can now write the Ka expression for the reaction. Ka is the equilibrium constant for acids and is defined as the product of the concentrations of the products divided by the concentration of the reactants: \(K_a = \dfrac{[H^+][X^-]}{[HX]}\) Plugging in the equilibrium concentrations from the ICE table: \(K_a = \dfrac{(0.075)(0.075)}{0.225}\) Finally, calculate Ka: \(K_a = \dfrac{0.005625}{0.225} = 0.025 \) So, the Ka value for the acid HX is 0.025.

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Most popular questions from this chapter

Hemoglobin (abbreviated Hb) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that are the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is pH- dependent. The relevant equilibrium reaction is $$ \mathrm{HbH}_{4}^{4+}(a q)+4 O_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q) $$ Use Le Châtelier's principle to answer the following. a. What form of hemoglobin, HbH \(_{4}^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4},\) is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood is decreased. How does this affect the oxygen-binding equilibrium? How does breathing into a paper bag help to counteract this effect? (See Exercise \(146 .\) ) c. When a person has suffered a cardiac arrest, injection of a sodium bicarbonate solution is given. Why is this necessary? (Hint: CO, blood levels increase during cardiac arrest.)

What are the major species present in 0.250\(M\) solutions of each of the following acids? Calculate the pH of each of these solutions. a. \(\mathrm{HOC}_{6} \mathrm{H}_{5}\) b. HCN

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? \(\left(K_{\mathrm{a}} \text { of HNCO }\right.\) \(=1.2 \times 10^{-4} . )\)

Would you expect \(\mathrm{Fe}^{3+}\) or \(\mathrm{Fe}^{2+}\) to be the stronger Lewis acid? Explain.

Sodium azide \(\left(\mathrm{NaN}_{3}\right)\) is sometimes added to water to kill bacteria. Calculate the concentration of all species in a \(0.010-M\) solution of \(\mathrm{NaN}_{3} .\) The \(K_{\mathrm{a}}\) value for hydrazoic acid (HN_) is \(1.9 \times 10^{-5} .\)
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