Mass of salt is equal to 218.8125 grams.
First we have to find Ka1 and Ka2
pKa1 = - log Ka1 so Ka1 = 0.059
pKa2 = - log Ka2 so Ka2 = 6.46 x 10⁻⁵
Looking at the values of equilibrium constants we can see that the first one is really big compared to second one. so, the pH will be affected mainly by the first ionization of the acid.
Oxalic acid is H₂C₂O₄
H₂C₂O₄ ⇄ H⁺ + HC₂O₄⁻
0.0356 M 0 0
0.0356 - x x x
Ka1 =
![\frac{[H^+][HC2O4^-]}{[H2C2O4]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BH%5E%2B%5D%5BHC2O4%5E-%5D%7D%7B%5BH2C2O4%5D%7D%20)
= x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6
Answer: The final pressure is 34.48kPa
Initial Pressure P1 = 55.16kPa
Initial Volume V1 = 0.500L
Final Pressure P2 = ?
Final Volume V2 = 0.800L
Boyle's law P1V1 = P2 V2
P2 = P1V1/V2
P2 = 55.16*0.5/0.8
P2 = 34.48kPa
C, potassium. Hope this helps.