Help me on B & C please ! (= The triprotic acid H3A has ionization constants of
ID: 879214 • Letter: H
Question
Help me on B & C please ! (= 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
Given a diprotic acid, H2A, with two ionization constants of Ka 3.06x 10-4 and Ka2 5.49x 10-2, calculate the pH and molar concentrations of H2A, HA-, and A for each of the solutions below (a) a 0.192 M solution of H2A [H,A] HA Number Number Number (b) a 0.192 M solution of NaHA [HA HA Number Number Number (c) a 0.192 M solution of Na2 [H,A] pH HA Number Number NumberExplanation / Answer
b-Ka1=3.06x10^-4, Ka2 = 5.4x10^-12
H2A<==>H+ + HA-
HA-<==> H+ + A2-
0.192-x x x
K1= [H+][HA-] / [H2A]
[H+]=[HA-] = 0.192M
pH= - log [H+] = - log 0.192=0.7166=0.72
[H2A]= [H+][HA-] / Ka1= 0.192 x 0.192 / 3.06x10^-4 = 0.01204x10^4 = 120.4M
Ka2= [H+][A2-] /[HA-]
5.49x10^-12= x^2/0.192-x
x is very small
5.49x10^-12 = x^2/0.192
x= [A2-]= 1.0267x10^-6M
c- [Na2A]=[H2A] = 0.192M
Ka1 = [H+][HA-]/[H2A]
Ka1= x^2/ 0.192-x
dissociation constant is very small so x is very small
Ka1=x^2/0.192
x= 0.7664x10^-2=[H+]= [HA-]
pH= -log [H+] = 2- 0.1155=1.8845
Ka2= [H+][A2-] /[HA-]
Ka2= y^2 / 0.7664x10^-2-y
y is very small
y2= Ka2x 0.7664x10^-2= 4.207x10^-14
y=[A2-]=2.05122 x10^-7M
ans-
Ka1= 8.4x10^-4, Ka2= 9.1x10^-7, Ka3= 9x10^-11
H3A<==> H+ + H2A-
[H2A-]= [H+] = 0.0320
pH=-log [H+]=-log 0.0320= 1.49485
H2A- <==> H+ + HA2-
0.032-x x x
Ka2= x^2/ 0.032-x [dissociation constant is very very small so x can be neglected]
x^2= 9.1x10^-7 x0.0320 = 0.2912x10^-7
x=[HA-] = 1.7064x10^-4M
HA2- <==> H+ + A3-
x-y y y
Ka3= y2/x-y [y is very small]
y^2= 9x10^-11 x1.7064x10^-4=
y=[A3-] = 12.39x10^-8M
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