Equilibrium Question 303
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 $
BETA
