Multiple Choice
Calculate the pH of a 0.400 M NaF solution, given that the Ka of HF is 6.4 x 10^-4.
17. Acid and Base Equilibrium
pH of Weak Bases
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Calculate the pH of a 0.021 M NaCN solution. Given that the Ka of HCN is 4.9 × 10^−10, what is the pH of the solution?
A
9.21
B
7.00
C
2.89
D
11.11
Verified step by step guidance1
Identify that NaCN is a salt that will dissociate in water to form Na+ and CN-. The CN- ion is the conjugate base of the weak acid HCN.
Recognize that CN- will undergo hydrolysis in water, reacting with water to form HCN and OH-. This reaction can be represented as: CN- + H2O ⇌ HCN + OH-.
Use the relationship between Ka and Kb to find the Kb for CN-. The formula is: Kb = Kw / Ka, where Kw is the ion-product constant of water (1.0 × 10^-14 at 25°C).
Set up the expression for the equilibrium constant Kb: Kb = [HCN][OH-] / [CN-]. Assume that the change in concentration of CN- is small, so [CN-] ≈ initial concentration.
Use the Kb expression to solve for [OH-], then calculate the pOH of the solution. Finally, convert pOH to pH using the relationship: pH + pOH = 14.
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