Multiple Choice
Pyridinium bromide (C5H5NHBr) is a strong electrolyte that dissociates completely into C5H5NH+ and Br-. A solution of pyridinium bromide has a pH of 2.92. Kb for pyridine is 1.7×10^-9. What is the molarity of the solution?
17. Acid and Base Equilibrium
pH of Strong Acids and Bases
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Join thousands of students who trust us to help them ace their exams!Watch the first videoMultiple Choice
What is the pH and [OH-] concentration of a solution if the [H3O+] is 4.0 x 10^-9 M?
A
pH = 7, [OH-] = 1.0 x 10^-7 M
B
pH = 9, [OH-] = 2.5 x 10^-6 M
C
pH = 5, [OH-] = 2.5 x 10^-10 M
D
pH = 3, [OH-] = 1.0 x 10^-11 M
Verified step by step guidance1
To find the pH of the solution, use the formula: \( \text{pH} = -\log[\text{H}_3\text{O}^+] \). Substitute the given \([\text{H}_3\text{O}^+] = 4.0 \times 10^{-9} \text{ M}\) into the formula.
Calculate the pH by evaluating the expression \( -\log(4.0 \times 10^{-9}) \). This will give you the pH of the solution.
To find the \([\text{OH}^-]\) concentration, use the water dissociation constant \(K_w = 1.0 \times 10^{-14} \text{ at 25°C}\), and the relationship \([\text{H}_3\text{O}^+][\text{OH}^-] = K_w\).
Rearrange the equation to solve for \([\text{OH}^-]\): \([\text{OH}^-] = \frac{K_w}{[\text{H}_3\text{O}^+]}\). Substitute \(K_w = 1.0 \times 10^{-14}\) and \([\text{H}_3\text{O}^+] = 4.0 \times 10^{-9} \text{ M}\) into the equation.
Calculate \([\text{OH}^-]\) by evaluating \( \frac{1.0 \times 10^{-14}}{4.0 \times 10^{-9}} \). This will give you the \([\text{OH}^-]\) concentration in the solution.
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