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Chapter 14: Problem 128

Calculate the pH of a \(0.10-M \mathrm{CoCl}_{3}\) solution. The \(K_{\mathrm{a}}\) value for \(\mathrm{Co}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is $1.0 \times 10^{-5} .$

Short Answer

Expert verified
The pH of the 0.10 M CoCl₃ solution is 3. We find this by first setting up an ICE table and using the given Kₐ value for Co(H₂O)₆³⁺ to calculate the H⁺ concentration, and finally applying the pH formula: \(pH = -\log_{10} [H^+] = -\log_{10} (1.0 \times 10^{-3})\).

Step by step solution

01

Write the ionization equilibrium

Write the ionization equilibrium for Co(H₂O)₆³⁺: \(Co(H_2 O)_6^{3+} \rightleftharpoons H^+ + Co(H_2 O)_5^{2+}\)
02

Set up an ICE table

Set up an Initial, Change, Equilibrium (ICE) table for the given equilibrium. The initial concentration of Co(H₂O)₆³⁺ is 0.10 M, while the initial concentration of H⁺ and Co(H₂O)₅²⁺ is 0. | | Co(H₂O)₆³⁺ | H⁺ | Co(H₂O)₅²⁺ | |-----|-----------|----|------------| | I | 0.10 | 0 | 0 | | C | -x | +x | +x | | E | 0.10 - x | x | x |
03

Write the expression for Kₐ

Write the expression for the Kₐ and plug in equilibrium concentrations from the ICE table: \( K_a = \frac{[H^+][Co(H_2 O)_5^{2+}]}{[Co(H_2 O)_6^{3+}]} = \frac{x * x}{0.10 - x} \)
04

Solve for x

We know that Kₐ is \(1.0 \times 10^{-5}\). Plug it into the expression and solve for x: \(1.0 \times 10^{-5} = \frac{x^2}{0.10 - x} \) As x is very small compared to 0.10, we can approximate it as: \(1.0 \times 10^{-5} = \frac{x^2}{0.10} \) Now, solve for x: \(x^2 = 1.0 \times 10^{-5} \times 0.10 \) \(x = \sqrt{1.0 \times 10^{-6}} \) \(x = 1.0 \times 10^{-3} \)
05

Calculate the pH

Now that we have found the concentration of H⁺ ions as \(1.0 \times 10^{-3} M\), we can calculate the pH of the solution using the pH formula: \(pH = -\log_{10} [H^+] = -\log_{10} (1.0 \times 10^{-3})\) The pH of the solution is 3.

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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 solution is prepared by dissolving 0.56 g benzoic acid $\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\( in enough water to make 1.0 \)\mathrm{L}$ of solution. Calculate $\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}\right],\left[\mathrm{H}^{+}\right]$ \(\left[\mathrm{OH}^{-}\right],\) and the pH of this solution.

Calculate the pH of an aqueous solution containing \(1.0 \times 10^{-2} M\) \(\mathrm{HCl}, 1.0 \times 10^{-2} \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4},\) and \(1.0 \times 10^{-2} \mathrm{M} \mathrm{HCN}\)

Calculate the \(\mathrm{pH}\) of a $0.050-M \mathrm{Al}\left(\mathrm{NO}_{3}\right)_{3}\( solution. The \)K_{\mathrm{a}}$ value for \(\mathrm{Al}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{3+}\) is $1.4 \times 10^{-5} .$

What mass of \(\mathrm{NaOH}(s)\) must be added to 1.0 \(\mathrm{L}\) of 0.050 \(\mathrm{M}\) \(\mathrm{NH}_{3}\) to ensure that the percent ionization of \(\mathrm{NH}_{3}\) is no greater than 0.0010\(\% ?\) Assume no volume change on addition of \(\mathrm{NaOH} .\)
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