25.0mL ofO.1 M NaOH is titrated with 0.1 MHCI. Calculate pH when i) 20mL.
ii) 24 mL of acid is added

Options

A. 12.0 , 11.30
B. 11.30, 12
C. 2.0, 2.70
D. 2.70, 2.0

Answer

A

Explanation

m mole of `NaOH = 25 xx 0.1 = 2.5`
(i) 20 mL of 0.1 M HCl is added.
m mol of HCl `= 20 xx 0.1 = 2`
m mol of NaOH left `= 2.5 - 2.0 = 0.5` m mol
Volume `= 25 + 20 = 45 mL, [overset(ɵ)(O)H] = ("0.5 m mol")/("45 mL") ~~ 0.01 = 10^(-2)M, pOH = 2, pH = 12`
(ii) 24 mL of 0.1 M HCl is added.
m mol of `HCl = 24 xx 0.1 = 2.4`, m mol of NaOH left `= 2.5 -2.4 = 0.1`
`[overset(ɵ)(O)H] = (0.1)/((25+24)) = 0.002 = 2 xx 10^(-3)M, pOH = -log[2 xx 10^(-3)] = -0.3 + 3 = 2.7`
`pH = 14 - 2.7 = 11.3`