CHEMISTRY PLS HELP

Question

For the diprotic weak acid H2A, 𝐾a1=3.5×10−6 and 𝐾a2=5.9×10−9.

What is the pH of a 0.0450 M solution of H2A? What are the equilibrium concentrations of H2A and A2− in this solution?

Step 3: Calculate the pH, [H2A], and [A2−] of the 0.0450 M solution of H2A at equilibrium

pH=

[H2A]=

[A2−]=

Answer 1

In the first dissociation of H2A:

molarity  H2A(aq)↔ (HA)^-(aq) + H^+(aq)

initial        0.045 m     0 m     0 m

change       -x        +x       +x

equilibrium  0.045-x       x        x

we can neglect X in [H2A] as it so small compared to the 0.045

so by substitution in Ka1 equation:

Ka1 = [HA][H] / [H2A]

3.5x10^-6 = X^2/0.045

X = √(3.5x10^-6)*(0.045)=

X= 3.968x10^-4 M

∴ [H2A] = 0.045 - 3.968x10^-4 = 0.0446 M

[HA] = 3.968x10^-4 M

[H] = 3.968x10^-4 M

the second dissociation of H2A: when ka2 = 5.9x10^-9

              HA-(aq)  ↔ A^2- (aq) + H+(aq)

at equilibrium 3.968x10^-4 M    y       3.968x10^-4 M

Ka2     = [H+][A^2-] / [HA]

5.9x10^-9 = Y(3.968x10^-4 M)/(3.968x10^-4 M)

∴y = 5.9x10^-9M

∴[A] = 5.9x10^-9M

PH= -㏒[H+]

  = -㏒(3.968x10^-4 M)= 3.401

[A]=5.9x10^-9M

[H2A] = 0.0446