Question

1) Calculate the concentration of H⁺ in a 0.550 M solution of ozalic acid. Conside only [H⁺] from the first ionization, K_a = 5.37 * 10^-2 .

2) Calculate the cocentration of H⁺ for the same solution, including the H⁺ ions resulting from the second ionization, K_a = 5.37 * 10-5

3) What is the expected pH for this solution?

4)Will the pH change by one unit if the solution is diluted to 0.0550 M?

Answer 1

Q1.

For acid:

H2C2O4 <-> H+ + HC2O4-

Ka1 = [H+][HC2O4-]/[HC2O4]

Ka1 = 5.37*10^-2

initially:

[H+]= 0

[HC2O4-] = 0

[HC2O4]= 0.55

in equilibrium

[H+]= 0+ x

[HC2O4-] = 0 + x

[HC2O4]= 0.55 - x

substitute in Ka

5.37*10^-2 = x*x/(0.55-x)

solve for x

x = 0.147

pH = -log(0.147) = 0.8326

Q2.

calculate H+ from second ionization:

[HC2O4-] = 0 + x = 0.147

so

HC2O4- <--> H+ + C2O4-2

KA2 =[H+][C2O4-2]/[HC2O4-]

initially:

[H+]= x

[C2O4-2]= 0

[HC2O4-] =  0.147

in equilibrium:

[H+]= x+y = 0.147+y

[C2O4-2]= 0+y

[HC2O4-] =  0.147-y

substitute:

5.37*10^-5 = (0.147+y)(y) /( 0.147-y)

solve for y

(5.37*10^-5)(0.147) - y(5.37*10^-5) = 0.147y + y^2

y^2 + (0.147+5.37*10^-5)y - (5.37*10^-5)(0.147)= 0

y^2 + 0.1470537 - 0.0000078939 =0

y = 5.3*10^-5

[H+]= x+y = 0.147+y = 0.147+5.3*10^-5 = 0.147053

pH = -log(0.147053) = 0.83252

Q3.

the expected pH is the 2nd ionization

that is, pH = 0.83252

Q4.

YES, pH changes if we dilute acid, sicne H+ is therefore diluted