Subscript in a pair of brackets
It's lone a little distinction (103 degrees versus 104 degrees in water), and I trust the standard rationalization is that since F is more electronegative than H, the electrons in the O-F bond invest more energy far from the O (and near the F) than the electrons in the O-H bond. That moves the powerful focal point of the unpleasant constrain between the bonding sets far from the O, and thus far from each other. So the shock between the bonding sets is marginally less, while the repugnance between the solitary matches on the O is the same - the outcome is the edge between the bonds is somewhat less.
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
What your question for number 3