Multiple Choice
What is the pH of the buffer prepared by mixing 200 mL of 0.15 M HNO2 (Ka = 4.0 x 10^-4) and 150 mL of 0.10 M sodium nitrite?
17. Acid and Base Equilibrium
pH of Weak Acids
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What is the pH of a 0.050 M solution of HNO2 (Ka = 4.6 × 10^-4)?
A
3.12
B
2.89
C
2.34
D
3.45
Verified step by step guidance1
Identify that HNO2 is a weak acid and will partially dissociate in water. The dissociation can be represented as: HNO2 ⇌ H+ + NO2−.
Write the expression for the acid dissociation constant (Ka) for HNO2: \( K_a = \frac{[H^+][NO_2^-]}{[HNO_2]} \).
Set up an ICE (Initial, Change, Equilibrium) table to determine the concentrations of the species at equilibrium. Initially, [HNO2] = 0.050 M, [H+] = 0, and [NO2−] = 0.
Assume that the change in concentration of HNO2 is 'x', so at equilibrium, [HNO2] = 0.050 - x, [H+] = x, and [NO2−] = x. Substitute these into the Ka expression: \( 4.6 \times 10^{-4} = \frac{x^2}{0.050 - x} \).
Assume x is small compared to 0.050 M, simplifying the equation to \( 4.6 \times 10^{-4} = \frac{x^2}{0.050} \). Solve for x to find [H+], then calculate pH using the formula \( \text{pH} = -\log[H^+] \).
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