Q31 / 220 · NTA Abhyas (NEET) 2020 · Chemistry · Ionic Equilibrium

Calculate the pH of a solution obtained by mixing 50 ml of 0.01 M Ba(OH)2 solution with 50 ml water.

  • A
    6
  • B
    10
  • C
    2
  • Correct answer
    D
    12

Solution

N1V1Ba(OH)2=N2V2mixture
[Normality of solution of Ba(OH)2=2× molarity, as its mol. wt = 2 x eq. wt as n- factor is 2 for it. ]
0.01×2×50=N2×(50+50) (as it is assumed as 100% ionised).
N2=0.01×100100=0.01 N
pOH=-log[OH-]=-log10-2=2log10
pOH=2
pH+pOH=14
pH+2=14
pH=12

Hence, the correct answer is option: 'D'.

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