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Chapter 13: Problem 43

The \(\mathrm{pH}\) of a \(0.129 \mathrm{M}\) solution of a weak acid, \(\mathrm{HB}\), is \(2.34\). What is \(K_{\mathrm{a}}\) for the weak acid?

Short Answer

Expert verified
Based on the given step by step solution, the short answer to the problem is: 1. Calculate the hydrogen ion concentration, [H+], using the formula [H+] = 10^(-pH) with pH = 2.34. 2. Set up an ICE table for the weak acid equilibrium reaction. 3. Write the Ka expression: \[K_a= \frac{[H+][B−]}{[HB]}\] 4. Substitute the equilibrium concentration values from the ICE table into the Ka expression and approximate: \[K_a= \frac{([H+] from \ step \ 1)(x)}{0.129}\] 5. Solve for Ka using the calculated value of [H+] from step 1.

Step by step solution

01

Convert the pH value to the hydrogen ion concentration

First, we will use the pH formula, pH = -log[H+], to calculate the hydrogen ion concentration. The formula can be rewritten as [H+] = 10^(-pH). Calculate [H+] for the given pH. [H+] = 10^(-2.34)
02

Use the ICE table to calculate the change in concentrations

Next, set up an ICE table for the weak acid equilibrium reaction, HB <=> H+ + B-. The initial concentrations are [HB]= 0.129 M, [H+]= calculated in step 1, and [B-]= 0. Initial: [HB]= 0.129 M, [H+]= [H+] from step 1, [B-]= 0 Change: [HB]= -x, [H+]= +x, [B-]= +x Equilibrium: [HB]= 0.129-x, [H+]= [H+] from step 1 + x, [B-]= x In the equilibrium, change in the concentration of H+ (x) is smaller compared to its initial concentration ([H+] from step 1), which allows us to make the assumption that [H+]=[H+] from step 1.
03

Write the Ka expression

The Ka expression for the given equilibrium is: \[K_a= \frac{[H+][B−]}{[HB]}\]
04

Substitute the equilibrium concentrations into the Ka expression

Substitute the equilibrium concentration values from the ICE table into the Ka expression and solve for Ka. \[K_a= \frac{([H+] from \ step \ 1)(x)}{0.129-x}\] We assume that x is very small compared to 0.129, so we can approximate this as: \[K_a= \frac{([H+] from \ step \ 1)(x)}{0.129}\] Referring back to the ICE table, x is equal to [B-] and also approximately equal to [H+] - [H+] from step 1. Therefore: \[K_a= \frac{([H+] from \ step \1)([H+]-[H+] from \ step \1)}{0.129}\]
05

Solve for Ka

Now solve the equation to find Ka for the weak acid. Replace [H+] from step 1 with the value calculated in step 1 and solve for Ka.

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Most popular questions from this chapter

For each of the following reactions, indicate the Brønsted-Lowry acids and bases. What are the conjugate acid/base pairs? (a) \(\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CN}^{-}(a q) \rightleftharpoons \mathrm{HCN}(a q)+\mathrm{H}_{2} \mathrm{O}\) (b) \(\mathrm{HNO}_{2}(a q)+\mathrm{OH}^{-}(a q) \rightleftharpoons \mathrm{NO}_{2}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}\) (c) \(\mathrm{HCHO}_{2}(a q)+\mathrm{H}_{2} \mathrm{O} \rightleftharpoons \mathrm{CHO}_{2}^{-}(a q)+\mathrm{H}_{3} \mathrm{O}^{+}(a q)\)

Phenol, once known as carbolic acid, \(\mathrm{HC}_{6} \mathrm{H}_{5} \mathrm{O}\), is a weak acid. It was one of the first antiseptics used by Lister. Its \(K_{\mathrm{a}}\) is \(1.1 \times 10^{-10} .\) A solution of phenol is prepared by dissolving \(14.5 \mathrm{~g}\) of phenol in enough water to make \(892 \mathrm{~mL}\) of solution. For this solution, calculate (a) \(\mathrm{pH}\) (b) \% ionization

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Which of the following is/are true about a \(0.10 \mathrm{M}\) solution of a strong acid, HY? (a) \(\left[\mathrm{Y}^{-}\right]=0.10 \mathrm{M}\) (b) \([\mathrm{HY}]=0.10 \mathrm{M}\) (c) \(\left[\mathrm{H}^{+}\right]=0.10 \mathrm{M}\) (d) \(\mathrm{pH}=1.0\) (c) \(\left[\mathrm{H}^{+}\right]+\left[\mathrm{Y}^{-}\right]=0.20 \mathrm{M}\)
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