Multiple Choice
What is the pH of a 0.050 M solution of HNO2 (Ka = 4.6 × 10⁻⁴)?
17. Acid and Base Equilibrium
pH of Weak Acids
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Consider a 1.0 M solution of a hypothetical weak acid HA whose pKa is 11. Calculate the pH of this solution.
A
pH = 11
B
pH = 1
C
pH = 7
D
pH = 5.5
Verified step by step guidance1
Start by understanding the relationship between pKa and the strength of an acid. The pKa value is a measure of the acid's dissociation in water. A higher pKa indicates a weaker acid.
Use the formula for the dissociation constant (Ka) of the weak acid: \( K_a = 10^{-pK_a} \). Substitute the given pKa value of 11 into this formula to find Ka.
Set up the equilibrium expression for the dissociation of the weak acid HA: \( HA \rightleftharpoons H^+ + A^- \). The initial concentration of HA is 1.0 M, and the change in concentration due to dissociation is represented by x.
Write the expression for Ka in terms of the equilibrium concentrations: \( K_a = \frac{[H^+][A^-]}{[HA]} \). Substitute the equilibrium concentrations: \( K_a = \frac{x^2}{1.0 - x} \).
Solve for x, which represents the concentration of \( H^+ \) ions. Once x is found, calculate the pH using the formula \( pH = -\log[H^+] \).
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