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 =
= x² / 0.0356 - x
x = 0.025 M
pH = - log [H⁺] = - log (0.025) = 1.6
The volume that will occupy at STP is calculated as follows
by use of ideal gas equation
that is PV=nRT where n is number of moles calculate number of moles
n= PV/RT
p=0.75 atm
V=6.0 L
R = 0.0821 L.atm/k.mol
T= 35 +273= 308k
n=?
n= (o.75 atm x 6.0 L)/( 0.0821 L.atm/k.mol x 308 k)= 0.178 moles
Agt STP 1 mole= 22.4 L what obout 0.178 moles
= 22.4 x0.178moles/ 1moles =3.98 L( answer C)
total pressure = ( 705 torr + 231.525torr) = 936.525torr