Consider the titration of a 26.0-mL sample of 0.185 M CH3NH2 (Kb = 4.4 × 10⁻⁴) with 0.155 M HBr. What is the initial pH of the CH3NH2 solution before any HBr is added?

• A. 9.85
• B. 7.00
• C. 11.22
• D. 2.78

Answer: (C)

Verified step by step guidance

1. First, identify that CH3NH2 is a weak base. To find the initial pH, we need to calculate the concentration of OH⁻ ions in the solution.
2. Use the base dissociation constant (Kb) for CH3NH2, which is given as 4.4 × 10⁻⁴. The equilibrium expression for the dissociation of CH3NH2 in water is: $K_b=\frac{\mathrm{[CH3NH3^+]}\mathrm{[OH^-]}}{\mathrm{[CH3NH2]}}$
3. Set up an ICE table (Initial, Change, Equilibrium) for the dissociation of CH3NH2. Initially, the concentration of CH3NH2 is 0.185 M, and the concentrations of CH3NH3⁺ and OH⁻ are 0 M.
4. Assume that x is the change in concentration for CH3NH3⁺ and OH⁻ at equilibrium. Therefore, the equilibrium concentrations are: $\mathrm{[CH3NH2]}=0.185-x$, $\mathrm{[CH3NH3^+]}=x$, $\mathrm{[OH^-]}=x$
5. Substitute these equilibrium concentrations into the Kb expression and solve for x, which represents [OH⁻]. Then, calculate the pOH using the formula: $\mathrm{pOH}=-\log \left(\mathrm{[OH^-]}\right)$. Finally, convert pOH to pH using: $\mathrm{pH}=14-\mathrm{pOH}$