# Figuring out the concentration of weak polyprotic acids in presence of strong acid

We have a $\rm{0.1 \ M}$ solution of $\rm{H_2S}$ with a $\rm{0.1 \ M}$ solution of $\rm{HCl}$. We need to figure out the concentration of $\rm{S^{2-}}$ ions at equilibrium.

$$\rm{H_2S \rightleftharpoons HS^{-} + H^+}$$

$$\rm{HS^- \rightleftharpoons S^{2-} + H^+}$$

Given $K_{a_1} = 9.1 \times 10^{-8}, K_{a_2} = 1.2\times 10^{-13}$.

I have no idea how to do this problem. I can't relate the $\rm{S^{2-}}$ concentration to the first equation. I first tried determining the $\rm{HS^-}$ concentration from the first equation, then putting that into the second one to obtain the required concentration, but then I realized that as the second reaction approaches equilibrium, the concentrations of the $\rm{HS^-}$ ions will change, which will then throw the first reaction out of whack.

I tried taking the required concentration as $y \ \rm{M}$ and then tried to establish a relationship, but the only thing I could get was that the concentration of the hydrogen ions will go up by $y$, which is of no use at all.

• You should probably be able to assume that the second equilibrium doesn't significantly shift the first. Any $\ce{HS^{-}}$ consumed to make $\ce{S^{2-}}$ will be almost completely regenerated to the previous amount, since there is a very large excess of $\ce{H2S}$. – Nicolau Saker Neto Feb 12 '14 at 11:13

## 1 Answer

The first step in solving this kind of problems is to write the reactions that take place in your solution, and the equations of the acidity constants :

$$\ce{H2S + H2O <=> HS^- + H3O^+}$$ $$K_{a_1}=\frac{[\ce{HS^-}][\ce{H3O^+}]}{[\ce{H_2S}]}$$

$$\ce{HS^- + H2O <=> S^{2-} + H3O^+}$$ $$K_{a_2}=\frac{[\ce{S^{2-}}][\ce{H3O^+}]}{[\ce{HS^-}]}$$ Since $\ce{HCl}$ is a strong acid ($pK_a < 0$), the dissociation equilibrium for it will be totally shifted to right: $$\ce{HCl + H2O -> H3O^+ + Cl^-}$$ And the dissociation of water, occuring in any solution: $$\ce{H2O + H2O <=> H3O^+ + HO^-}$$ $$K_w = [\ce{H3O^+}][\ce{HO^-}]$$ *The concentration of water is not taken into account because it remains approximately constant as it is much bigger than the concentration of the $\ce{H2S}$ dissolved. It is also right to write the equations as you did above, considering the dissociation of the acid, but it's better to use the hydronium ion, $\ce{H3O^+}$, as we know $\ce{H^+}$ is quite unstable on its own. (In some books $[\ce{H^+}]$ is used to ease writing)

As you saw, you cannot calculate the concentrations at equilibrium (written in square brackets, i.e. $[\ce{HS^-}]$) considering only the first reaction (unless you consider the shift of the second reaction doesn't affect significantly the first, as Nicolau said, and so we use the analytical concentration of a compound (in this case $\ce{H2S}$ and $\ce{HCl}$) which means the specific concentration of the solute when the solution is prepared ($0,1M$ for both acids, if these are the concentrations in the mixture of acids solution and not the initial concentration of each acid solution). It is usually denoted $c_{\ce{H2S}}$ and $c_{\ce{HCl}}$ respectively.

Looking at the above reactions, we can write the formula for the analytical concentrations : $$c_{\ce{H2S}} = [\ce{H2S}] + [\ce{HS^-}] + [\ce{S^{2-}}]$$ and $$c_{\ce{HCl}} = [\ce{Cl^-}]$$ As you might notice no $\ce{H3O^+}$ concentration appeared in any of these formulas, because it depends on both $\ce{HCl}$ and $\ce{H2S}$ concentrations.

However, this is not enough to solve our problem, and so the electroneutrality principle is used, stating that the concentration of the negative ions must be equal to the concentration of the positive ones, and in our case may be written as: $$[\ce{HS^-}] + 2[\ce{S^{2-}}] + [\ce{Cl-}] + [\ce{HO^-}] = [\ce{H3O^+}]$$ *The concentration of $\ce{S^{2-}}$ has the coefficient $2$ because of the two negative charges, unlike the other ions, which have only one negative or positive charge and don't need any.

Using the equations of the analytical concentrations in the electroneutrality formula, we get this : $$c_{\ce{H2S}} - [\ce{H2S}] + [\ce{S^{2-}}] + c_{\ce{HCl}} + [\ce{HO^-}] = [\ce{H3O^+}]$$

And, using also the acidity constants and the water constant : $$c_{\ce{H2S}}-\frac{[\ce{H3O^+}][\ce{HS^-}]}{K_{a_1}}+\frac{K_{a_2}[\ce{HS^-}]}{[\ce{H3O^+}]} + c_{\ce{HCl}} + \frac{K_w}{[\ce{H3O^+}]} = [\ce{H3O^+}]$$

The equation has 2 unknown concentrations ($[\ce{H3O^+}]$ and $[\ce{HS^-}]$) and you may approximate the concentration of $\ce{H3O^+}$ to $c_{\ce{HCl}}$ since the acidity constants for $\ce{H2S}$ are very small: $$[\ce{H3O^+}]=c_{\ce{HCl}}$$ So, the concentration of $\ce{HS^-}$ can be calculated: $$[\ce{HS^-}]=\frac{c_{\ce{H2S}} + \frac{K_w}{c_{\ce{HCl}}}}{\frac{c_{\ce{HCl}}}{K_{a_1}} + \frac{K_{a_2}}{c_{\ce{HCl}}}}$$

However, since $\frac{K_w}{c_{\ce{HCl}}} << c_{\ce{H2S}}$ and $\frac{K_{a_2}}{c_{\ce{HCl}}} << \frac{c_{\ce{HCl}}}{K_{a_1}}$, the above equation can be approximated to: $$[\ce{HS^-}] = \frac{c_{\ce{H2S}}K_{a_1}}{c_{\ce{HCl}}}$$ The approximations lead us to a simple formula, which could be denoted directly from the first equilibrium, but without understanding exactly why. From this the concentration of $\ce{S^{2-}}$ can be calculated easily, just like you did first. Even though some would just ignore the second equilibrium for $\ce{H2S}$, that is NOT the way to solve analytical chemistry problems. First you consider all of the equilibriums and processes occuring in a solution and then (if possible) approximate, and not the reverse.