he triprotic acid H3A has ionization constants of Ka1 = 8.4× 10–4, Ka2 = 9.1× 10
ID: 879108 • Letter: H
Question
he 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
I have one more attempt, Please help me (:
The triprotic acid HyA has ionization constants of Kr1-84x 10-4, Ka2 = 9. 1x 10-7, and Ka,-9.0x10-11. Mapd Calculate the following values for a 0.0320 M solution of NaH2A Number Number H,A [H-J= 111.48 HjA Calculate the following values for a 0.0320 M solution of Na2HA Number Number HA H2.01 Previous Check Answer 0 Next Exit Hint Consider H2A to be the intermediate form of a diprotic acid, surrounded by HsA and HA2. Consider HA to be the intermediate form of a diprotic acid, surrounded by HA and A3. Use the equations for [H associated with the intermediate form in a diprotic system.Explanation / Answer
NaH2A is a salt will hydrolyze as,
H2A- + H2O à H3A+ + OH-
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 à 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
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.