What is the [S^-2] in a saturated solution (.1 M) of H2S, in which the pH had been adjusted to 5.00 by the addition of NaOH? For H2S, Ka1= 1.1X10^-7 and Ka2=1.0X10^-14
To find the concentration of [S^-2] in the saturated solution of H2S, we need to consider the dissociation of H2S in water and the equilibrium expressions for its dissociation reactions.
The dissociation of H2S can be represented by the following equilibrium reactions:
1) H2S ⇌ H+ + HS^- (Ka1 = 1.1X10^-7)
2) HS^- ⇌ H+ + S^-2 (Ka2 = 1.0X10^-14)
First, let's determine the concentration of [H+] in the saturated solution. We are given that the pH is adjusted to 5.00 by the addition of NaOH. Since NaOH is a strong base, it fully dissociates to produce OH^- ions. Therefore, we can determine the concentration of [OH^-] by converting the pH to pOH and then calculating the concentration of [OH^-].
pOH = 14 - pH
pOH = 14 - 5.00
pOH = 9.00
[OH^-] = 10^(-pOH)
[OH^-] = 10^(-9.00)
[OH^-] = 1.0 x 10^(-9) M
Since water is neutral, the concentration of [H+] is equal to the concentration of [OH^-]. Therefore, [H+] also has a concentration of 1.0 x 10^(-9) M.
Now we can use the equilibrium expressions to find the concentrations of [HS^-] and [S^-2].
For the reaction H2S ⇌ H+ + HS^-, we have:
Ka1 = [H+][HS^-] / [H2S]
Plugging in the values we know:
1.1X10^-7 = (1.0X10^-9)([HS^-]) / (0.1)
Solving for [HS^-]:
[HS^-] = (1.1X10^-7)(0.1) / (1.0X10^-9)
[HS^-] = 1.1X10^-8 M
Now we can use the second equilibrium expression for the reaction HS^- ⇌ H+ + S^-2:
Ka2 = [H+][S^-2] / [HS^-]
Plugging in the values we know:
1.0X10^-14 = (1.0X10^-9)([S^-2]) / (1.1X10^-8)
Solving for [S^-2]:
[S^-2] = (1.1X10^-8)(1.0X10^-14) / (1.0X10^-9)
[S^-2] = 1.1X10^-13 M
Therefore, the concentration of [S^-2] in the saturated solution (0.1 M) of H2S, in which the pH had been adjusted to 5.00, is 1.1 x 10^-13 M.
Those k values don't look right to me but I assume you copied them as is from your worksheet.
(H^+)^2(S^-2)/(H2S) = k1k2
You have only one unknown, the (S^-2). Substitute and solve for that.