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H2S dissociated in 2 steps. K1= 10^-7 and K2=10^-12.9. The total amount of sulfur bearing species...

H2S dissociated in 2 steps. K1= 10^-7 and K2=10^-12.9. The total amount of sulfur bearing species is 0.01mol/L. Calculate HS-, H2S, and S2- at pH=2

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Answer #1

At pH = 2, the following equilibrium exists.

H2S HS-

According to Henderson-Hasselbulch equation: pH = pKa1 + Log([HS-]/[H2S])

i.e. 2 = 7 + Log([HS-]/[H2S]) (pKa1 = -Log Ka1)

i.e. Log([H2S]/[HS-]) = 7-2 = 5

i.e. [H2S]/[HS-] = 105

i.e. [H2S] = 105 * [HS-] ............. Equation 1

According to given data: [H2S] + [HS-] = 10-2 M ................. Equaiton 2

From equaitons 1 and 2, you can write as follows.

105 * [HS-] + [HS-] = 10-2

i.e. 105 * [HS-] = 10-2 (compared to 105, 1 can be easily neglected)

i.e. [HS-] = 9.9999*10-8 M

Substitute the above value in equation 2, you will get [H2S] = 9.9999*10-3 M

And [S2-] = 0

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