1.

What is [H^(+)] of a solution that is 0.01 M in HCN and 0.02 Min NaCN (K_(a) for HCN = 6.2 xx 10^(-10))

Answer»

`3.1 XX 10^(10)`
`6.2 xx 10^(5)`
`6.2 xx 10^(-10)`
`3.1 xx 10^(-10)`

Solution :`K_(a) = ([H^(+)][CN^(-)])/([HCN^(-)]) rArr 6.2 xx 10^(-10) = ([H^(+)] [0.02])/([0.01])`
`[H^(+)] = (6.2 xx 10^(-10) xx 0.01)/(0.02) = 3.1 xx 10^(-10)`.


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