Precipitates and pH?

Question

These two questions are really bugging me. Anyone have any ideas?

A 0.05 M solution of hydrogen sulfide (H2S) contains 0.15 M nickel chloride NiCl2, and 0.35 M mercury nitrate Hg(NO3)2. What pH is required to precipitate the maximum amount of HgS but none of the NiS?

Even before the industrial age, rainwater was discovered to be slightly acidic from dissolved CO2. What is the pH of a bottle of rainwater collected in 1863 if the solubility of CO2 in pure water is 88 mL CO2 gas in 100 mL H2O. The Ka1 of H2CO3 is equal to 4.5 x 10^-7. the volume percent of CO2 in air is 0.033%.

Answer 1

You need to provide Kps values and Ka1, Ka2 for H2S.

Both sulfides are of the type MS with a reaction for dissolving MS <=> M+2 + S-2 and
Ksp = [M+2][S-2]

so the minimum concentration of [S-2] needed for each one to start precipitating is

[S-2]= Ksp/[[M+2]

The concentration you are looking for is that for [S-2]=KspNiS/[NiS] since at that concentration you have a saturated NiS solution and you can't go any further because NiS will start precipitating.

Once you find [S-2] you can find [H+] based on the following:

H2S is a diprotic acid so if you really want to be accurate it is proven that for a diprotic acid (for derivation of the equation look at http://www.chembuddy.com/?left=ph-calcul... where Ka12= Ka1*Ka2)

[S-2] = Ka1*Ka2*Ca / ([H+]^2+Ka1[H+] + Ka1*Ka2)

You have calculated [S-2], Ca=0.05 and they should give you Ka1, Ka2 so you substitute and then solve the quadratic for [H+].

I don't see why they give you the percent in the air.
They should have also provided the temperature. Assuming 25 C, then T=298 and PV=nRT =>
n=PV/RT= 1*0.088/(0.082*298) =3.6*10^-3 mole CO2 in 100 mL water => C(CO2) =0.036 mole/L

If we assume that all of the CO2 that dissolves in water becomes H2CO3 (in practice this is not the case, but you are not giving the equilibrium constant), so Cacid= 0.036 M
They give you only the 1st dissociation constant so apparently you can neglect the second. Therefore

.. .. .. .. .. .. H2CO3 <=> HCO3(-) + H+
Initial .. .. .. .. 0.036
Dissociate .. .. x
Produce .. .. .. .. .. .. .. .. .. .. x .. .. .. .. x
At Equil .. 0.036-x .. .. .. .. ..x .. .. .. .. x

Ka1= x^2/(0.036-x)
Let's assume that 0.036 >> x so that 0.036-x =0.036
Then
Ka1=x^2/0.036 => x=Squareroot (0.036*Ka1) =.
x =SQRT(0.036*4.5*10^-7) = 1.27*10^-4 our assumption is fair enough. If you want solve the quadratic for higher accuracy.
pH= -log(1.27*10^-4) =3.896 =3.90

Answer 2

For the first one, start with the expression for Ksp of NiS. You should be able to calculate the [S2-] at which NiS just begins to precipitate. Then, you should be able to use the Ka expression for HS-, and calculate the pH which will produce that concentration of S2-.

Don't ya just love Analytical Chem?

I'll let someone else try the second one...

Answer 3

Cu(OH)2 = Cu ++ + 2 OH- pH = 8.6, so pOH = 14 - 8.6 = 5.4 [OH-] = antilog (-pOH) = antilog (-5.4) = 4.0 Exp -6 Ksp = [Cu++][OH-]^2 = 4.8 Exp -20 [Cu++](4.0 Exp -6)^2 = 4.8 Exp -20 [Cu++](a million.6 Exp -11) = 4.8 Exp -20 [Cu++] = 4.8 Exp -20 / (a million.6 Exp -11) [Cu++] = 3.0 Exp -9 answer: [Cu++] = 3.0 Exp -9 M