NCERT Class XI Chemistry Equilibrium Solutions

Question : 48

Total: 73

Assuming complete dissociation, calculate the pH of the following solutions:
(a) 0.003 M HCl
(b) 0.005 M NaOH
(c) 0.002 M HBr
(d) 0.002 M KOH

Solution:

(a) HCl + aq → H++Cl–
∴ [H+] = [HCl] = 3 × 10−3 M, pH = - log (3×10−3) = 2.52
(b) NaOH + aq → Na++OH−
∴ [OH−] = 5 × 10−3M
∴ [H+] =

10−4

5×10−3

= 2 × 10−12M , pH = - log (2 × 10−12) = 11.70
(c) HBr + aq → H++Br−
∴ [H+] = 2 × 10−3 M, pH = - log (2×10−3) = 2.70
(d) KOH + aq → K++OH−
∴ [OH−] = 2 × 10−3M , [H+] =

10−14

2×10−3

= 5 × 10−12
pH = - log (5×10−12) = 11.30

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