A 20.0-mLmL sample of a 0.200 MHBrMHBr solution is titrated with a 0.200 MNaOHMNaOH solution. Calculate the pHpH of the solution after the following volumes of base have been added.

This question is incomplete, the complete question is;

A 20.0-mLmL sample of a 0.200 MHBr solution is titrated with a 0.200 MNaOH solution. Calculate the pH of the solution after the following volumes of base have been added;

a. 14.0 mL

b. 19.7 mL

c. 20.2 mL

d. 35.0 mL

a)  PH = 1.452

b)  PH = 2.820

c) pH = 10.998

d) pH = 12.737

Explanation:

a)

Given that;

Volume of HBr Va  = 20.0 mL

Normality of HBr ( Na ) = 0.200 N

Normality of NaOH ( Nb ) = 0.200 N

Volume of NaOH ( Vb ) = 14.0 mL

m.eq acid = NaVa = 0.200×20.0 = 4.0

m.eq of base = NbVb = 0.200 × 14.0 = 2.8

m.eq acid > m.eq base , so solution is acidic

[ H+ ] = (NaVa - NbVb) /( Va + Vb)

[ H+ ] = (4.0 - 2.8 ) /( 20.0 + 14.0) = 0.03529

pH = - log [ H+ ]

pH = - log 0.03529

PH = 1.452

b)

Volume of base ( Vb ) = 19.7

m.eq base = NbVb = 0.200 × 19.7 = 3.94

m.eq acid > m.eq base , so solution is acidic

[ H+ ] = (NaVa - NbVb) / Va + Vb

[ H+ ] = (4.0 - 3.94 ) / (20.0 + 19.7) = 0.00151

pH = - log [ H+ ]

pH = - log 0.00151

PH = 2.820

c)

Volume of base ( Vb ) = 20.2

m.eq base = NbVb = 0.200 × 20.2= 4.04

m.eq base > m.eq acid , so solution is basic

[ OH- ] = (NbVb - NaVa) / Va + Vb

[ OH- ] = (4.04 - 4.0) / (20.0 + 20.2) = 0.000995

pOH = - log [ OH- ]

pOH = - log 0.000995

POH = 3.002

pH = 14 - pOH

pH = 14 - 3.002 = 10.998

d)

Volume of base ( Vb ) = 35.0

m.eq base = NbVb = 0.200 × 35.0= 7

m.eq base > m.eq acid , so solution is basic

[ OH- ] = (NbVb - NaVa) / Va + Vb

[ OH- ] = (7.0 - 4.0) / (20.0 + 35.0) = 0.0545

pOH = - log [ OH- ]

pOH = - log 0.0545

POH = 1.263

pH = 14 - pOH

pH = 14 - 1.263 = 12.737

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