Answer:
C
Explanation:
Because i just had to do this on my assignment and got 100
From the chemical equation given:
H2SO4+2KOH--->K2SO4+2H2O
the two reactants, H2SO4 and KOH, are in 1:2 stoichiometric ratio.
No. of moles of KOH = 2* no. of moles of H2SO4
=2*0.1*0.033
The concentration of KOH = no. of moles / volume
=2*0.1*0.033/0.05
=0.132M
According to the reaction equation:
Hf(aq) + H2O (l)↔ H3O+ (aq) + F(aq)
initial 1 m 0 0
-X +X +X
(1-x) X X
We assumed that [H3O+] = X
so. at eqilibrium:
Kc = [H3O] [ F] / [Hf] when we have Kc so by substitution:
3.5x10^-4 = X^2 / (1-X)
∴X^2 = 3.5x10^-4 - 3.5x10^-4 X
X^2 + 3.5x10^-4 X - 3.5x10^-4= Zero
by solving this equation:
(X-1.9x10^-2)(X + 1.9x10^-2) = Zero
∴X = 1.9x10^-2
∴ the equilibrium constatnt of H3O = 1.9x10^-2 M
