Question

9 wo solutions consisting of 257.0 mL of 0. 5 M lead(IV) nitrate and 515,0 mL of 0.6 M sodium sulfate are mixed. What is the final concentration of the aqueous sulfate ion? Hint: Don't forget to account for the precipitation reaction! Pb(NOs)4(aq)+2Na2SO,(aq) Ph(SO)(0)+4NANO;(aq) dimt (A) 4.6736 M Sou (B) 0. 133 81 M (C) 0.3608 M (D) 6. 7358 x 10-2 M l325

Answer 1

Let us find out limiting reactant from the given data
for that we need moles of both reactants

Lead nitrate Volume =257 ml= 0.257 L
molarity of lead nitrate= 0.5 M
# moles of lead nitrate =molarity of lead nitrate *volume of lead nitrate
moles of lead nitrate present = 0.257 L*0.5 M= 0.1285 mole**

volume of sodium sulphate = 515 ml = 0.515 L
molarity of sodium sulphate= 0.6M
moles of sodium sulphate present =0.515L*0.6 M= 0.309 moles**

Now let us find out limiting reagent
Since lead nitrate has fewer moles
it will act as limiting reagent and it wiil consume completely.
And some moles of Na2So4 will remain in the solution .We need to
know how much moles of Na2SO4 is there in the solution
for that lets us calculate how much moles of Na2SO4 is consumed
to form products
Pb(NO3)4 + 2Na2SO4 --> Pb[SO4]2 + 4 NaNO3

Is the ballanced equation given >From this
we can write mole ratio

Pb(NO3)4 :Na2SO4 = 1:2
This means 1 mole Pb(NO3)4 needs 2 moles of Na2SO4

but here
moles of lead nitrate present = 0.1285 mole**
so
0.1285 mole Pb(NO3)4 needs [2*0.1285] moles of Na2SO4
number of moles of Na2SO4 = 0.257
Thus 0.1285 moles Pb(NO3)4 reacts with 0.257 moles Na2So4 to from products***

Thu moles of Na2SO4 consumed = 0.257
So moles of Na2SO4 remain in the solution =
moles of sodium sulphate initally present -moles of sodium sulphate consumed
= 0.309 - 0.257 = 0.052 moles**

now let us see how Na2So4 break up in aq.solution
Na2SO4 ---> 2Na+ + SO42-
this swos 1 mole Na2SO4 produce 1 mole SO42- in aq solution
So 0.052 moles Na2SO4 produces 0.052 moles SO42- ions

THus moles of SO42- in solution = 0.052 mols
Total volume of solution = 0.257 L + 0.515 L = 0.772 L
Thus molarity of SO4 2- = moles of SO42- in solution/total volume of solution
=6.7358 x10^-2 M
option D
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