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NH4 OH hArr NH(4)^(+) + OH^(-) ,Kb = 2 x...

`NH_4 OH hArr NH_(4)^(+) + OH^(-) ,K_b = 2 xx 10^(-5)` .If one litre of a solution contains 0.1 mole each of ammonia and ammonium sulphate, calculate the pH of buffer solution

Text Solution

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`k_b` of `NH_4 OH = 2 xx 10^(-5)`
hence `pK_a ` of `NH_4 OH =- log k_b = 5 - log 2`
` [NH_3] =0.1 mol, [NH_(4)^(+)]=2[ (NH_4)_2 SO_4]=0.2 mol`
` pOH = pK_b + log ""([NH_(4)^(+) ])/([NH_3]) = 5- log 2 + log (0.2 )/( 0.1 ) = 5.0`
` pH =-pOH =14-5=9`
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