1.

Calculate pH of a basic soluion OH^(-) as (i) 10^(-2) M ,and(ii) 10^(-8) M

Answer»

Solution :(i) `10^(-2) " M " OH^(-)`
` OH^(-) = 10^(-2) , POH = -log [OH^(-)] = 2 `
` pH = 14-2 =2`
(ii)` 10^(-6) " M " OH^(-)`
In presenceof `10^(-8) " M " OH^(-) ` , WATER dissociates as
`{:(H_(2)O,hArr,H^(+),+,OH^(-)),((C-x),,x,,(10^(-8)+x)):}`
`K_(W) = [H^(+)] [OH]= (10^(-8)+x) x = 10^(-14)`
Which reduces to quadratic equation
` x = (-10^(-8)pmsqrt(10^(-8)+4xx10^(-14)))/2`
` :.x = 0.95 xx10^(-7)` mol/L
`[OH]= (10^(-8) +x) = 10^(-8) +0.95 xx10^(-7)`
` =1.05 xx10^(-7)`
`pOH = - log [OH] = -log (1.05 xx10^(-7)) = 6.9788`
` pH = 14 - pOH = 14 - 6.9788 = 7.0212`


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