How pH can relate to the Ksp

Precipitation mechanism of sulfides

-pH and sulfide ion

The sulfide ion is the base resulting from the double dissociation of hydrogen sulphide, a weak diprotic acid:

k and q

For the double dissociation of hydrogen sulphide:

$\ Dpi {120} H_2S \ rightleftharpoons 2H + ^ + S ^ {2-}$

the overall equation for this equilibrium is achieved by combining the two previous ones:

$\ Dpi {120} K_ {tot} = K_1 \ cdot K_2 = \ frac {[H ^ {+}] [HS ^ {-}]} {[H_2S]} \ cdot \ frac {[H ^ {+}] [S ^ {2 -}]} {[HS ^ {-}]} = \ frac {[H ^ {+}] ^ {2} [S ^ {2 -}]} {[H_2S]} = 1, 1 \ cdot 10 ^ {- 22}$

From these can be derived the concentration of the sulfide ion has a function of pH:

$[S ^ {2-}] = \ frac {K_ {tot} \ cdot [H_2S]} {[H ^ +] ^ {2}}$ Hydrogen sulphide is a gas at room temperature. In a saturated solution at a pressure of 1 atmosphere (laboratory standard condition), we have [H₂S] equal to 0.1 M.

We got therefore a perfect straight relationship between pH and concentration of sulfide ions:

$[S ^ {2-}] = \ frac {1.1 \ cdot 10 ^ {- 23}} {[H ^ +] ^ 2}$

This report is crucial, for example, in the approach to the analysis of group 2 cations. It doens't matter how many sulfide ions are freed by Na₂S . The concentration of S^2-, as you just saw, is determined by pH.

From this equation we notice that:

for high pH values, ie for low concentrations of H^+, the concentration of free sulfide ions is high.
Conversely, for low values of pH, or high concentration of H⁺, the concentration of S^2- available it's very low. Here are the numerical examples:

We always perform the calculation with S^2- as unknown because the pH characterizes our aqueous solution and determines the concentration of sulfide which will actually be present in solution. We can not say the opposite (is not the sulfide who determines the pH) simply because the amount of sulfide added (wheter we are working properly) is very small compared to the volume of the solution.

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