Question

25 mL of 0.10 M Pb (NO 3 ) 2 is mixed with 50 - mL of 0.090 M Na 2 SO 4 .

a) Write an equation for a possible precipitation reaction.

b) Under the conditions specified will a precipitate from

Answer 1

(a) The precipitation reaction is as follows :

Pb(NO3)2 + Na2SO4 ----> PbSO4 + 2NaNO3

(b) Pb(NO3)2 ---> Pb ^2+ + 2NO3^-

Na2SO4 ----> 2Na^+ + SO4^2-

2 moles of na+ is produced from 1 mole of Na2SO4.

1 mole of Pb ^2+ is produced from 1 mole of PbSO4.

Moles of PbSO4 present = 0.1 *25mL/1000mL = 0.0025 moles.

Moles of Na2SO4 present = 0.09 *50mL/1000 = 0.0045 moles

So, total molarity of Pb^2+ in the mixture = moles of Pb^2+ *1000mL/(25+50)= 0.025 *1000/75 = 0.033 M

Total molarity of SO4^2- in the mixture = moles of SO4^2- *1000mL/(25+50) = 0.0045 *1000/75 = 0.06 M

[Pb][SO4] = 0.033 *0.06 =0.00198

Ksp of PbSO4 =2.53 *10^-8

[Pb][SO4]>>> Ksp. So, a precipitate will form.