2 For the diprotic weak acid H2A, Ka1 2 7 10 6 and Ka2 8 8 10 9 ...

H2A HA H I 005 0 0 C x x x E 005x x x ka1 HHAH2A ....

2. For the diprotic weak acid H2A, Ka1 = 2.7 Ã— 10-6 and Ka2 = 8.8...

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2. For the diprotic weak acid H2A, Ka1 = 2.7 Ã— 10-6 and Ka2 =8.8 Ã— 10-9. What is the pH of a 0.0500 M solution of H2A? What arethe equilibrium concentrations of H2A and A2â€“ in this solution?

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Â Â Â Â Â Â Â Â Â Â Â Â H2A ---------> HA^-Â Â + H⁺

^Â Â IÂ Â Â Â Â Â Â Â Â 0.05Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â 0

CÂ Â Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â +x

Â Â EÂ Â Â Â Â Â 0.05-xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â +xÂ Â Â Â Â Â Â +x

Â Â Â Â Â Â Â Â ka1Â Â = [H+][HA-]/[H2A]

Â Â Â Â Â Â Â Â 2.7*10^-6Â Â = x*x/0.05-x

Â Â Â Â Â Â Â Â 2.7*10^-6 *(0.05-x) = x²

Â Â Â Â Â Â Â Â Â Â Â Â x = 0.000366

Â Â Â Â Â Â Â Â [H+] = x = 0.000366M

Â Â Â Â Â Â Â Â [HA-]Â Â Â = x = 0.000366M

Â Â Â Â Â Â Â HA^- -------> A^2- + H⁺

Â Â Ka2 = [A^2-][H⁺]/[HA^-]

8.8*10^-9Â Â = [A^2-]*0.000366/0.000366

Â Â [A^2-]Â Â = 8.8*10^-9 M

PH = -log[H+]

Â Â Â Â Â Â Â Â = -log0.000366Â Â = 3.4365

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2. For the diprotic weak acid H2A, Ka1 = 2.7 Ã— 10-6 and Ka2 = 8.8...

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60.1K

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Question

Chemistry

2. For the diprotic weak acid H2A, Ka1 = 2.7 Ã— 10-6 and Ka2 =8.8 Ã— 10-9. What is the pH of a 0.0500 M solution of H2A? What arethe equilibrium concentrations of H2A and A2â€“ in this solution?

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