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edgar chambukeni mzuni
Answers (1)
[HX] = .250moles/(655/1000L) =.38M HXH+ + x- I .38 0 0 C -X +x +x E .38-x x x pH = 3.44 to find H+ -log[H+] = pH = H= 3.63x 10^-4 Ka = [H+][x-]/[HX] = [3.63x10^-4]^2/3.8x10^-1 = 3.5x10^-9
