Question

# In a titration of a 100.0mL 1.00M HA weak acid solution with 1.00M NaOH, what is...

In a titration of a 100.0mL 1.00M HA weak acid solution with 1.00M NaOH, what is the pH of the solution after the addition of 135 mL of NaOH? Ka = 1.80 x 10-5 for HA. (3 significant figures)

The weak acid will react with NaOH and will get neutralized by it as

HA + NaOH --> Na+ + A- + H2O

The equivalence point will be when moles of acid will become equal to moles of base

Moles of acid present = molarity of acid X volume of acid in Litres

Moles of acid = 1 X 0.1 = 0.1 moles

Moles of base added = Molarity of base X volume of base in litres

Moles of base = 1 X 0.135 = 0.135 moles

So the base will completely neutralize the acid

The moles of base left = 0.135 - 0.1 = 0.035

Total volume = 100 + 135 = 235mL = 0.235 L

So concentration of base left = 0.035 / 0.235 = 0.149 molar

As the base is strong base, it will dissociate completely and will give equal concentration of OH-

[OH-] = 0.149

pOH = -log[OH-] = -log[0.149] = 0.827

so pH = 14 - 0.827 = 13.173 = 13.2

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