JIPMER 2019 Evening Papers Online Practice

(2)
N1V1−N2V2 = NV
base acid
NaOH (HBr)
50 × 0.2 – 50 × 0.1 = N × [100]
[OH‌−] = N =

5

100

[OH‌−] = 5 × 10‌−2
pOH = – log [OH‌−]
= – log (5 × 10‌−2 )
= 2 – log 5
= 2 – 0.7
pOH = 1.3
pH + pOH = 14
pH = 14 – 1.3
pH = 12.7