# chemistry

Calculate the pH of a solution obtained by mixing 50 mL of NH3 0.1 M (Kb = 1.8•10–5 mol/L) and 25 mL of HCl 0.2 M.

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1. HCl + NH3 ==> NH4Cl
mmols HCl = 5
mmols NH3 = 5
So you have a solution of 5 mmols NH4Cl iin 75 mL and M = mmols/mL = 5/75 = 0.0667.

The pH of the salt is determined by the hydrolysis of the NH4^+
.....NH4^+ + H2O ==> H3O^+ + NH3
I...0.0667............0.......0
C....-x...............x.......x
E..0.0667-x...........x.......x

Ka for NH4^+ = (Kw/Kb for NH3) = (x)(x)/(0.0667-x). Solve for x = (H3O^+) and convert to pH.

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