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Question: pH of Ba(OH)2 solution is 12. Its solubility product is...

pH of Ba(OH)2 solution is 12. Its solubility product is

A

10−6M310 ^ { - 6 } \mathrm { M } ^ { 3 }

B

4×10−6M34 \times 10 ^ { - 6 } \mathrm { M } ^ { 3 }

C

0.5×10−7M30.5 \times 10 ^ { - 7 } \mathrm { M } ^ { 3 }

D

5×10−7M35 \times 10 ^ { - 7 } \mathrm { M } ^ { 3 }

Answer

5×10−7M35 \times 10 ^ { - 7 } \mathrm { M } ^ { 3 }

Explanation

Solution

Since pH=12pOH=14−12=2\mathrm { pH } = 12 \mathrm { pOH } = 14 - 12 = 2

∴[Ba++]=10−22M\therefore \left[ \mathrm { Ba } ^ { + + } \right] = \frac { 10 ^ { - 2 } } { 2 } \mathrm { M } ∴Ksp=[Ba++][OH−]2=(10−22)2=5×10−7M3\therefore \mathrm { Ksp } = \left[ \mathrm { Ba } ^ { + + } \right] \left[ \mathrm { OH } ^ { - } \right] ^ { 2 } = \left( \frac { 10 ^ { - 2 } } { 2 } \right) ^ { 2 } = 5 \times 10 ^ { - 7 } \mathrm { M } ^ { 3 }

We know Ba(OH)2Ba+++2OH−\mathrm { Ba } ( \mathrm { OH } ) _ { 2 } \quad \mathrm { Ba } ^ { + + } + 2 \mathrm { OH } ^ { - }