Tuesday, March 29, 2011

Calculate the pH of a solution by mixing 40.0 mL of a 0.02 M HCl solution with 200.0 mL of 0.20 M HCN solution. Assume volumes to be additive.Ka...

HCl is a strong acid and HCN is a weak acid.


Given


Volume of HCl = 40ml = 0.04L


Molarity of HCl = 0.02M


Volume of HCN = 200ml = 0.2L


Molarity of HCN = 0.2M


First we have to find individual moles.


Moles = Molarity * volume.


Moles of HCl = 0.02 * 0.04 = 0.0008


Moles of HCN = 0.2 * 0.2 = 0.04


Total volume = 0.04 + 0.2 = 0.24


Total moles = 0.0008 + 0.04 = 0.0408


Total concentration = moles / volume = 0.0408/0.24 = 0.17


Ka = (x^2)/(0.17-x)


1.0x10^-10 = (x^2)/(0.17-x)


1.0x10^-10 = (x^2)/(0.17)


X^2 = (1.0x10^-10)(0.17)


X = 4.1 * 10^-6


Ph = -log(H+)


Ph = -log(4.1 * 10^-6)


Ph = 5.3.

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