Answer to Question #170595 in General Chemistry for Surendra

Question #170595

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


1
Expert's answer
2021-03-15T09:06:17-0400

"K_a=\\frac{K_w} {K_b} =\\frac{1\u00d710^{-14}}{1.7\u00d710^{-9}}"


"=5.88\u00d710^{-6}=\\frac{x^{2}} {0.025-x}=\\frac{x^{2}}{0.025}"


"x=1.21\u00d710^{-2}M=[H^{+}]"


"pH=-log(1.21\u00d710^{-2})=1.92"


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