A 50.0 mL solution of 0.140 M KOH is titrated with 0.280 M HC1. Calculate the pH of the solution ...

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a) For 0.00 ml

There is only KOH present in the solution, which is a strong base, hence it will dissociate fully into the solution

b) After adding 5.00 ml of HCl

Moles of HCl added = 5.00/1000 * 0.280 = 0.0014 moles

Moles of KOH present = 50.0/1000 * 0.140 = 0.007 moles

Moles of KOH left after reaction = 0.007 - 0.0014 = 0.0056 moles

Volume of solution = 50.0 + 5.0 = 55.0

Molarity of [OH-] = Number of moles/Volume of solution (in L) = 0.0056/55.0 * 1000 = 0.1018

pOH = -log[0.1018] = 0.99

pH = 14 - pOH = 14 - 0.99 = 13.01

c) After adding 12.50 ml of HCl

Moles of HCl added = 12.50/1000 * 0.280 = 0.0035 moles

Moles of KOH present = 50.0/1000 * 0.140 = 0.007 moles

Moles of KOH left after reaction = 0.007 - 0.0035 = 0.0035 moles

Volume of solution = 50.0 + 12.50 = 62.50

Molarity of [OH-] = Number of moles/Volume of solution (in L) = 0.0035/62.50 * 1000 = 0.056

pOH = -log[0.056] = 1.25

pH = 14 - pOH = 14 - 1.25 = 12.75

d)

After adding 19.0 ml of HCl

Moles of HCl added = 19.0/1000 * 0.280 = 0.00532 moles

Moles of KOH present = 50.0/1000 * 0.140 = 0.007 moles

Moles of KOH left after reaction = 0.007 - 0.00532 = 0.00168 moles

Volume of solution = 50.0 + 19.0 = 69.0

Molarity of [OH-] = Number of moles/Volume of solution (in L) = 0.00168/69.0 * 1000 = 0.0243

pOH = -log[0.0243] = 1.62

pH = 14 - pOH = 14 - 1.62 = 12.38