Answer to Question #266960 in General Chemistry for Paulina

Question #266960

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

1
Expert's answer
2021-11-17T05:01:01-0500

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

moles of NH3=0.2×0.5=0.1NH_3=0.2×0.5=0.1

pOH=14pHpOH=14−pH

pOH=149.44=4.56pOH=14−9.44=4.56

pKb=log[Kb]=log[1.78×105]=4.74pK_b ​=−log[K_b ​]=−log[1.78×10^{−5} ]=4.74

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

log[base][salt]=4.744.56=0.18log\frac{ [base]}{[salt]} ​=4.74−4.56=0.18

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


[base]=[base] = moles of NH3=0.1NH_3 ​=0.1


[salt]=0.11.51=0.07[salt]=\frac{ 0.1}{1.51} ​=0.07

Moles of (NH4)+=0.07(NH_4 ​)^+ =0.07

(NH4)2SO42NH4++(SO4)2(NH_4 ​)_2 ​SO_4 ​⇌2NH_4 ​^+ +(SO_4 ​)^{−2}


moles of (NH4)2SO4=0.072=0.035(NH_4 ​)_2 ​SO_4 ​=\frac{0.07}{2}=0.035


amount of (NH4)2SO4=0.035×132.14=4.6249g(NH4 ​)2 ​SO_4 ​=0.035×132.14=4.6249g



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