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Chapter 14: Problem 67

Calculate the concentration of all species present and the pH of a \(0.020-M\) HF solution.

Short Answer

Expert verified
The concentration of all species present in the 0.020-M HF solution are approximately: [HF] = 0.020-x M, [H+] = 0.0026 M, [F-] = 0.0026 M, and the pH of the solution is approximately 2.59.

Step by step solution

01

Write the reaction for the dissociation of the weak acid HF

In this step, write the balanced equation for the dissociation of the weak acid, HF, into its ions: \[HF(aq) \leftrightarrow H^+(aq) + F^-(aq)\]
02

Write the Ka expression for HF

Write the acid dissociation constant (Ka) expression for HF dissociation: \[K_a = \frac{[H^+][F^-]}{[HF]}\]
03

Create an equilibrium table

Express the initial concentrations, changes, and equilibrium concentrations as a table using the initial concentration of HF (0.020 M) and variables x: | | HF | H+ | F- | |:---:|:-----:|:--:|:--:| |Initial|0.020 M| 0 | 0 | |Change |-x | +x | +x | |Equilibrium|0.020-x| x | x |
04

Substitute the variables in the Ka expression and solve for x

Substitute the variables of the equilibrium expressions into the Ka expression and use the given Ka value for HF(6.8 × 10^(-4)). \[(6.8 \times 10^{-4}) = \frac{x^2}{0.020-x}\] Assume x is very small compared to 0.020, hence we can ignore the x in the denominator. \[(6.8 \times 10^{-4}) \approx \frac{x^2}{0.020}\] Now, solve for x (H+ concentration): \[x \approx \sqrt{(6.8 \times 10^{-4})(0.020)}\] \[x \approx 0.0026\: \text{M}\]
05

Calculate the pH

Use the concentration of H+ (x) to calculate the pH: \[pH = -\log{[H^+]}\] \[pH \approx -\log{(0.0026)}\] \[pH \approx 2.59\] Therefore, the concentration of all species present are approximately: [HF] = 0.020-x M, [H+] = 0.0026 M, [F-] = 0.0026 M, and the pH of the 0.020 M HF solution is approximately 2.59.

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Most popular questions from this chapter

When determining the pH of a weak acid solution, sometimes the 5\(\%\) rule can be applied to simplify the math. At what \(K_{\mathrm{a}}\) values will a \(1.0-M\) solution of a weak acid follow the 5\(\%\) rule?

a. The principal equilibrium in a solution of \(\mathrm{NaHCO}_{3}\) is $$ \mathrm{HCO}_{3}^{-}(a q)+\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}_{3}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ Calculate the value of the equilibrium constant for this reaction. b. At equilibrium, what is the relationship between $\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]\( and \)\left[\mathrm{CO}_{3}^{2-}\right] ?$ c. Using the equilibrium $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons 2 \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) $$ derive an expression for the pH of the solution in terms of \(K_{\mathrm{a}_{1}}\) and \(K_{\mathrm{a}_{2}}\) using the result from part b. d. What is the pH of a solution of \(\mathrm{NaHCO}_{3} ?\)

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Calculate the percentage of pyridine $\left(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\right)\( that forms pyridinium ion, \)\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+},\( in a \)0.10-M$ aqueous solution of pyridine \(\left(K_{\mathrm{b}}=1.7 \times 10^{-9}\right)\)
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