What is the pH of the resulting solution if 45 mL of 0.432 mol L^-1 aqueous methylamine,CH₃NH₂,is added to 15 mL of 0.234 mol L^-1 HCl(aq) ?
Assume that the volumes of the solutions are additive.K_a = 2.70 × 10^-11for CH₃NH₃⁺.

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Quizplus · Accepted Answer

The reaction between methylamine (a weak base) and HCl (a strong acid) results in a neutralization reaction, forming water and a salt. Given the volumes and molarities, we first calculate the moles of each reactant:- Moles of CH3NH2 = 0.045 L * 0.432 mol/L = 0.01944 mol- Moles of HCl = 0.015 L * 0.234 mol/L = 0.00351 molSince methylamine is in excess (0.01944 mol - 0.00351 mol = 0.01593 mol remaining), we use the formula for the pH of a basic solution:$$ \text{pH} = 14 + \log\left(\frac{K_b \times [\text{base remaining}]}{0.060 L}\right) $$Given $K_b = 2.70 \times 10^{-11}$, and the total volume is 0.060 L (45 mL + 15 mL), we find:$$ \text{pH} = 14 + \log\left(\frac{2.70 \times 10^{-11} \times 0.01593}{0.060}\right) $$This calculation yields a pH greater than 7, indicating a basic solution. Among the provided options, only option D (11.23) is plausible for a basic solution resulting from an excess of a weak base over a strong acid.

What Is the PH of the Resulting Solution If 45