Question

what is the pH of 25 mL of 0.15M H2C2O4 mixed with 25mL of 0.30M NaOH

(Ka1= 5.9x10^-2, Ka2= 6.4x10^-5)

Answer 1

The reactions occuring are

Initial concentration of [H2C2O4] =

25 ml of 0.15 M = 3.75×10-3 M

ka1 = 5.9×10-2

At equilibrium

We get

x3 +(0.059)x -( 2.2125×10-4)

Solving we get

x = 3.749×10-3

[H+]1 = x2 = (3.749×10-3) 2= 1.405×10-5 M

for NaOH

Concentration =25 ml of 0.30 M

= 0.025(0.30) = 7.5×10-3 M

ka2 = 6.4×10-5

At equilibrium

x2 +(6.4×10-5) x -(4.8×10-7) = 0

Solving we get

x = 6.615×10-4 M

[OH-] = 6.615×10-4 M

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

[H+] = 10-14/(6.615×10-4)

[H+]2 = 1.5117×10-11 M

[H+]= [H+]1 +[H+]2

Total volume = 25+25 = 50 mL

[H+] =( (1.5117×10-11) +(1.405×10-5)) 0.050 = 7.025×10-7 M

pH = -log[H+] = -log(7.025×10-7) =6.1533

pH of mixture = 6.153

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