Equilibrium Question 432

Question: $ 50ml $ water is added to a $ 50ml $ solution of $ Ba{{(OH)}_2} $ of strength $ 0.01M. $ The $ pH $ value of the resulting solution will be [MP PMT 1999]

Options:

A) 8

B) 10

C) 12

D) 6

Show Answer

Answer:

Correct Answer: C

Solution:

0.01 M $ Ba{{(OH)}_2}=0.02NBa{{(OH)}_2} $

$ N_1V_1=N_2V_2 $

$ [0.02N]\times [50ml]=N_2\times 100ml $

$ N_2=\frac{0.02\times 50}{100} $

$ ={10^{-2}}N $ ; $ [O{H^{-}}]={10^{-2}} $ N $ pOH=2 $ or $ pH=12 $



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