Answer to Question #266960 in General Chemistry for Paulina

Question

How many grams of (NH4)2SO4 (fw = 132.14 g/mol) should be added to 425 mL of a 0.258 M solution of NH3 to achieve a buffer solution with a pH of 9.44? Kb for NH3 = 1.8 x 10-5

Answer 1

Molarity of NH3 solution = moles of NH3/ volume of solution

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moles of $NH_3=0.2×0.5=0.1$

$pOH=14−pH$

$pOH=14−9.44=4.56$

$pK_b ​=−log[K_b ​]=−log[1.78×10^{−5} ]=4.74$

$pOH=pK_b ​−log\frac{ [base]}{[salt]}$

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$log\frac{ [base]}{[salt]} ​=4.74−4.56=0.18$

$\frac{[base]}{[salt]} ​=antilog[0.18]=1.51$

$[base] =$ moles of $NH_3 ​=0.1$

$[salt]=\frac{ 0.1}{1.51} ​=0.07$

Moles of $(NH_4 ​)^+ =0.07$

$(NH_4 ​)_2 ​SO_4 ​⇌2NH_4 ​^+ +(SO_4 ​)^{−2}$

moles of $(NH_4 ​)_2 ​SO_4 ​=\frac{0.07}{2}=0.035$

amount of $(NH4 ​)2 ​SO_4 ​=0.035×132.14=4.6249g$