A buffer solution contains 1 mol of ammonium | vedclass

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(D) The buffer is a basic buffer consisting of a weak base $NH_4OH$ and its salt $(NH_4)_2SO_4$.For a basic buffer,the $pOH$ is given by the Henderson

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$8.7$

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(D) The buffer is a basic buffer consisting of a weak base $NH_4OH$ and its salt $(NH_4)_2SO_4$.For a basic buffer,the $pOH$ is given by the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.Given: $n_{(NH_4)_2SO_4} = 1 \ mol$,so $[NH_4^+] = 2 	imes 1 = 2 \ mol$ (since one mole of salt provides two moles of ammonium ions).$n_{NH_4OH} = 1 \ mol$.$pK_b = -\log(10^{-5}) = 5$.$pOH = 5 + \log \frac{2}{1} = 5 + 0.3 = 5.3$.Since $pH + pOH = 14$,$pH = 14 - 5.3 = 8.7$.

$A$ buffer solution contains $1 \ mol$ of ammonium sulphate and $1 \ mol$ of $NH_4OH$ $(K_b = 10^{-5})$. The $pH$ of the solution will be:

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