Question

A 50.0 mL solution of 0.140 M KOH is titrated with 0.280 M HC1. Calculate the pH of the solution after the addition of each of the given amounts of HCI 0.00 mL pH = 11.84 5.00 mL pH = 12.5 mL pH = 12.74 19.0 mL pH = 12.38 24.0 mL pH = 11.56 25.0 mL pH = 7 31.0mL pH = 1.7

Answer 1

Solved the first four parts as per HOMEWORKLIB RULES, Looks like your remaining answers are correct. In case you need any more explanation, post it in comments section. Rate the answer positive if found helpful

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