Equilibrium Question 871
Question: In which case change in pH is maximum-
Options:
A) 1 mL of pH = 2 is diluted to 100 mL
B) 0.01 mol of $ NaOH $ is added to 100 mL of 0.01 M $ NaOH $ solution
C) 100 mL of $ H_2O $ is added to 900 mL of $ {10^{-6}} $ M $ HCl $
D) 100 mL of pH = 2 solution is mixed with 100 mL of pH = 12
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Answer:
Correct Answer: D
Solution:
$ pH $ before $ pH $ after Changes [a] 2 4 2 [b] 12 13.04 1.04 [c] 6 6.7 0.7 [d] 2 7 5 Explanation: [a] $ pH=2.[{H^{+}}]={10^{-2}}M $
$ M_1V_1=M_2V_2 $
$ {10^{-2}}\times 1=M_2\times 100 $
$ M_2={10^{-4}} $
$ \therefore $ $pH=4 $
[b] $ [ O{H^{-}} ]={10^{-2}} $
$ pOH=2 $
$ pH=12 $ 100 of 0.01 M $ NaOH $
contains 0.001 mol $ NaOH $ added - 0.01 mol Total moles = 0.011 mol
$ \therefore $ ${{[ O{H^{-1}} ]}_{final}}=0.011\times 10=0.11M $
$ pOH=0.96 $
pH = 13.04 [c] $ [ HCl ]={10^{-6}}M $
$ [ {H^{+}} ]={10^{-6}}M;pH=6 $
After dilution $ [ HCl ]={10^{-7}}M=[ {H^{+}} ] $
$ [ {H^{+}} ]in,H_2O={10^{-7}}M $
Total $ [ {H^{+}} ]={10^{-7}}+{10^{-7}}=2\times {10^{-7}} $
$ \therefore $ $pH=7-\log 2=6.7 $
[d] $ ,pH=2,,[ {H^{+}} ]={10^{-2}}M $
$ pH=12,,[ {H^{+}} ]={10^{-12}}M $
$ \therefore $ $,[ O{H^{-}} ]={10^{-2}}M $
BETA
