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0.001 mole of strong electrolyte Zn(OH)2...

0.001 mole of strong electrolyte Zn`(OH)_2` is present in 200 mL of an aqueous solution. The pH of this solution is

A

2

B

4

C

12

D

10

Text Solution

Verified by Experts

The correct Answer is:
C
`Zn(OH)_2 hArr Zn^(2+) + 2OH^(-)`
`[OH^-] = 0.001 xx2xx 1000/200`
`[OH^-] =1 xx 10^(-2) mol L^(-1)`
`POH =- log [OH^(-) ]= log (1 xx 10^(-2) ) = + 2 log 10`
`pOH =2, pH +POH = 14`
`pH +2=14 implies pH = 12`
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