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
Answer:
a) PH = 1.452
b) PH = 2.820
c) pH = 10.998
d) pH = 12.737
Step-by-step 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