0.10 moles of a diprotic acid, H2S, is dissolved in 250 mL of water(so concentra

Question

0.10 moles of a diprotic acid, H2S, is dissolved in 250 mL of water(so concentration = 0.10mol/0.25L=0.4M). The Ka1 of this acid is1.0 x 10E-5 and Ka2 is 1.0 x 10E-10. What is the concentration ofthe di-anion, [S^2-], in this solution (in M)?

a) 0.40
b) 2.0 x 10E-3
c) 1.0 x 10E-5
d) 4.0 x 10E-6
e) 1.0 x 10E-10

I think that the answer is e), but I am unsure. Please willsomeone assist me. I think that the procedure is asfollows:
1) from the deprotonation equilibrium of the acid (H2A), determinethe concentrations of the conjugate base (HA-) and H30+
2) Find the concentrations of A2- from the second deprotonationequilibrium (that of HA-) by substituting the concentrations ofH3O+ and HA- from step 1 into the expression for Ka2.

Explanation / Answer

0.10 moles of a diprotic acid, H2S, is dissolved in 250 mL of water(so concentra

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