Answer to Question #170595 in General Chemistry for Surendra

Question

Calculate the pH of a 0.025 M solution of CH.NHCI (Show your work 2 pts). C5H5NHCl

Answer 1

$K_a=\frac{K_w} {K_b} =\frac{1×10^{-14}}{1.7×10^{-9}}$

$=5.88×10^{-6}=\frac{x^{2}} {0.025-x}=\frac{x^{2}}{0.025}$

$x=1.21×10^{-2}M=[H^{+}]$

$pH=-log(1.21×10^{-2})=1.92$