1
Question
If pH of a saturated solution of Ba(OH)2 is 12, the value of its K(sp) is:
If pH of a saturated solution of Ba(OH)2 is 12, the value of its K(sp) is:
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Solution
The correct option is A 5.00×10−7
Let us consider the problem.
pH+pOH=14
pOH=14−12=2(givenpH=12)
pOH=−log[OH−]pOH=10−pOH=10−2........1
Ba(OH)2→Ba+2+2OH−
At equlibrium-Ba+2=xandOH−=2x
Since,2x=10−2asequation1,
Therefore ,
x=10−22=0.5×10−2KSP=[Ba+2]×[OH−]2=[0.5×10−2][10−2]2=0.5×10−6=5×10−7
Let us consider the problem.
pH+pOH=14
pOH=14−12=2(givenpH=12)
pOH=−log[OH−]
pOH=10−pOH=10−2........1
Ba(OH)2→Ba+2+2OH−
At equlibrium-
Ba+2=xandOH−=2x
Since,
2x=10−2asequation1,
Therefore ,
x=10−22=0.5×10−2
x=10−22=0.5×10−2
KSP=[Ba+2]×[OH−]2=[0.5×10−2][10−2]2=0.5×10−6=5×10−7
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