The triprotic acid H3A has ionization constants of Ka1 = 8.4× 10–4, Ka2 = 9.1× 1

Question

The triprotic acid H3A has ionization constants of Ka1 = 8.4× 10–4, Ka2 = 9.1× 10–7, and Ka3 = 9.0× 10–11. Calculate the following values for a 0.0320 M solution of NaH2A

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sapling learning The triprotic acid HyA has ionization constants of Kat-8.4x10-4, Ka2 9.1x10-7, and Ku3 = 90x 10-11. Calculate the following values for a 0.0320 M solution of NaH2A Number Number H2A [H.J= 111.48 H,A Calculate the following values for a 0.0320 M solution of Na2HA. Number Number H2.01 H,A

Explanation / Answer

NaH2A is a salt will hydrolyze as,

H2A- + H2O --> H3A+ + OH-

hydrolysis constant = KH = [H3A+][OH-]/[H2A-]

The concentration of hydroxide ion can be calculated as

KH = Kw/Ka = 1 x 10^-14/8.4 x 10^-4 = 1.19 x 10^-11

                H2A- + H2O           -->          H3A+ + OH-

Initial          0.032                                   0         0

Change        -x                                       x         x

Equilibrium   0.032-x                                x     x

So

1.19 x 10^-11 = x^2/(0.032 - x)

x <<1

so x^2 = 1.19 X 10^-11 X 0.032 = 0.38 X 10^-12

x = [H3A+] = [OH-] = 6.166x 10^-7 M

[H+] X [OH-] = 10^-14

[H+] = 1.62 x 10^-8 M

The decrease in concentration of H2A- will be

H2A- = 0.032 - 6.16 x 10^-7 = 0.031999 M

So

[H2A-] /[H3A] = 51862.723

Now,

The salt H2A- dissociates as,

H2A- à HA2- + H+

Ka2 = [HA2-][H+]/[H2A-]

let x be the amount dissociated and [H+] from previous calculation then,

9.1 x 10^-7 = 6.16 x 10^-7 x [HA2-]/[H2A-]

[HA2-]/[H2A-] = 1.477

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