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The triprotic acid H3A has ionization constants of Ka1 = 1.6× 10–3, Ka2 = 2.4× 10–9,...

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Question

Chemistry

The triprotic acid H3A has ionization constants of Ka1 = 1.6×10–3, Ka2 = 2.4× 10–9, and Ka3 = 6.4× 10–11.

Calculate the following values for a 0.0800 M solution ofNaH2A.

[H+] = M

[H2A-]/[H3A] =   

Calculate the following values for a 0.0800M solution ofNa2HA.

[H+] = M

[HA2-]/[H2A-] =

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Hint:

Consider H2A– to be the intermediate form of a diprotic acid,surrounded by H3A and HA2–. Consider HA2– to be the intermediateform of a diprotic acid, surrounded by HA– and A3–. Use theequations for [H ] associated with the intermediate form in adiprotic system.

Answer & Explanation Solved by verified expert
4.2 Ratings (505 Votes)
Na2HA H2O Na NaHA OH I    See Answer
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