[chrissy]

So here's a question i got 5/10 for and i'm not sure where i went wrong. I redid it and got different values so i would be very grateful if somebody took a careful look at it and told me if it seems to make sense :

One way lead(II) ions can be efficiently taken out of an aqueous solution before waste disposal is by precipitating them out with sodium hydroxide.
a) We have 15 L of aqueous waste with an estimated 4.0 x 10^-3 mol/L concentration of lead (II) ions. A technician adds 3.5g of sodium hydroxide to it. What mass of lead (II) ions could be recovered from this mixture?

What i did :

eq: Pb+2(aq)+ 2 NaOH (aq) --> Pb(OH)_2(s) + 2Na+(aq)

Find moles: M= moles/ L so moles= M*L= 4*10^-3(15)= 0.06 moles of Pb+2

then i found the moles of NaOH from the eq: g/mm= moles so 3.5/38.008= 0.092 moles of NaOH
then i set up my IRE table (that some call IRF but its just initial reactant final or equilibrium)



               Pb+2(aq)   +   2 NaOH(aq)   ----->   Pb(OH)_2 (s)   +    2Na+(aq)

1             0.06                 0.092                         0                         0

R            -0.046              - 0.092                  +0.046                   +0.092

E           0.014                    0                          0.046                   0.092

so in the end I have 0.046 moles of Pb(OH)_2 but were looking for ions of lead (II) only
molar mass of lead (II) = 207.2 g so using g/mm= moles we get 0.046(207.2) = g = 9.5 g of Pb+2

b) what would be the concentration of the excess ion left in the mixture?

ion left is 2Na+  eq: moles/ volume = []

0.092/15L = 6.13*10^-3 M




thanks a million for any comments!

[cliverlong]

1) Essentially I agree with your method but for M_r of NaOH I use the value 23+16+1 = 40 leading to 0.0875 mole NaOH and 9.06g Pb²⁺ react

2) The final mixture doesn't contain only Na⁺, it also contains unreacted Pb²⁺

Clive