1.

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

Answer»

2
4
12
10

Solution :`Zn(OH)_2hArrZn^(2+) + 2OH^(-)`
`[OH^-] = 0.001xx2xx 1000/200`
`[OH^-]=1 XX 10^(-2)mol L^(-1)`
`POH=- log[OH^(-) ]= log(1 xx 10^(-2) )= + 2 LOG10`
`pOH=2, pH+POH= 14`
`pH +2=14 impliespH = 12`


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