Multiple Choice
At what pH is the side chain of lysine 35% ionized?
17. Acid and Base Equilibrium
pH of Weak Acids
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What is the pH of a 0.115 M hydrogen chromate ion (HCrO₄⁻) solution, given that the Ka of HCrO₄⁻ is 3.2 x 10⁻⁷?
A
6.89
B
5.67
C
3.45
D
4.12
Verified step by step guidance1
Start by writing the dissociation equation for hydrogen chromate ion (HCrO₄⁻) in water: HCrO₄⁻ ⇌ H⁺ + CrO₄²⁻.
Use the expression for the acid dissociation constant (Ka) to set up the equilibrium expression: Ka = [H⁺][CrO₄²⁻] / [HCrO₄⁻].
Assume that the initial concentration of HCrO₄⁻ is 0.115 M and that the change in concentration due to dissociation is 'x'. Therefore, at equilibrium, [H⁺] = x, [CrO₄²⁻] = x, and [HCrO₄⁻] = 0.115 - x.
Substitute these equilibrium concentrations into the Ka expression: 3.2 x 10⁻⁷ = (x)(x) / (0.115 - x).
Solve the quadratic equation for 'x', which represents the concentration of H⁺ ions. Once 'x' is found, calculate the pH using the formula: pH = -log₁₀[H⁺].
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