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Chapter 14: Problem 164

Calculate the pH of a \(0.010-M\) solution of iodic acid (HIO $_{3}, K_{\mathrm{a}}\( \)=0.17 )$

Short Answer

Expert verified
The pH of a 0.010-M solution of iodic acid (HIO3) with \(K_a = 0.17\) is approximately 1.38.

Step by step solution

01

Write the chemical equation

Begin by writing the chemical equation for the ionization of iodic acid in water: \[HIO_3 (aq) + H_2O (l) \rightleftharpoons H_3O^+(aq) + IO_3^-(aq)\]
02

Write the equilibrium expression

Next, write the equilibrium expression (Ka) for the ionization of iodic acid in water: \[K_a = \frac{[H_3O^+][IO_3^-]}{[HIO_3]}\]
03

Define the equilibrium concentrations

Let x be the equilibrium concentration of H3O⁺ and IO3⁻ ions. The initial concentration of iodic acid is 0.010 M, so the equilibrium concentrations can be defined as: - [H3O⁺] = x - [IO3⁻] = x - [HIO3] = 0.010 - x
04

Substitute equilibrium concentrations into the Ka expression

Substitute the equilibrium concentrations in the Ka expression: \[0.17 = \frac{x^2}{0.010 - x}\]
05

Solve the quadratic equation for x

Since Ka is significantly larger than x, we can approximate the equation by neglecting x in the denominator: \[0.17 = \frac{x^2}{0.010}\] Now, solve for x: \[x^2 = 0.17 \times 0.010\] \[x^2 = 0.0017\] \[x = \sqrt{0.0017}\] \[x = 0.0412\]
06

Calculate the pH of the solution

Now that we have the concentration of H3O⁺ ions, we can calculate the pH of the solution using the formula: pH = -log[H3O⁺] \[pH = -log(0.0412)\] \[pH \approx 1.38\] The pH of a 0.010-M solution of iodic acid (HIO3) is approximately 1.38.

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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.)

Making use of the assumptions we ordinarily make in calculating the \(\mathrm{pH}\) of an aqueous solution of a weak acid, calculate the pH of a \(1.0 \times 10^{-6}-\mathrm{M}\) solution of hypobromous acid \(\left(\mathrm{HBrO}, K_{\mathrm{a}}=2 \times 10^{-9}\right) .\) What is wrong with your answer? Why is it wrong? Without trying to solve the problem, explain what has to be included to solve the problem correctly.

Calculate the percent dissociation of the acid in each of the following solutions. a. 0.50\(M\) acetic acid b. 0.050\(M\) acetic acid c. 0.0050\(M\) acetic acid d. Use Le Châtelier's principle to explain why percent dissociation increases as the concentration of a weak acid decreases. e. Even though the percent dissociation increases from solutions a to \(c,\) the \(\left[\mathrm{H}^{+}\right]\) decreases. Explain.

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?

Calculate the percent dissociation for a \(0.22-M\) solution of chlorous acid \(\left(\mathrm{HClO}_{2}, K_{\mathrm{a}}=0.012\right)\)
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