1.

Calculate pOH of 0.001 M NH_(4)OH. When it is 1% dissociated in the solution

Answer»

5
2.96
9.04
11.04

SOLUTION :`[OH^(-)]` in `NH_(4) Oh` solution `= C.alpha`
`RARR 0.001 xx (1)/(100) = 1 xx 10^(-5), pOH = -log[OH^(-)]`
`rArr log[1 xx 10^(-5)], pOH = 5`.


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