Question

Consider the titration of 20.0 mL of 0.100M HCl with 0.200 M NaOH solution. Calculate the pH after the addition of the following volumes of sodium hydroxide solution.

A) 0.00 mL

B) 12.00 mL

C) 20.00 mL

D) 25.00 mL

Answer 1

this is strong acid + strong base

then

initial mmol of acid = MV = 20*0.1 = 2

a)

[H+] = [HCl] = 0.1

pH = -log(0.1) = 1

b)

mmol of base = MV = 0.2*12 = 2.4

HCl + NaOH = H2O + NaCl

mmol of base left = 2.4 - 2 = 0.4

[NaOH] = mmol/V = 0.4 / (20+12) = 0.0125

pH = 14- pOH = 14 + log(OH) = 14+log(0.0125) = 12.09

c)

mmol of base = MV = 0.2*20= 4

HCl + NaOH = H2O + NaCl

mmol of base left = 4- 2 = 2

[NaOH] = mmol/V = 2/ (20+20) = 0.05

pH = 14- pOH = 14 + log(OH) = 14+log(0.05) = 12.69

d)

mmol of base = MV = 0.2*25= 5

HCl + NaOH = H2O + NaCl

mmol of base left = 5- 2 = 3

[NaOH] = mmol/V = 3/ (20+25) = 0.0666

pH = 14- pOH = 14 + log(OH) = 14+log(0.0666) = 12.82