Answer to Question #105911 in Chemistry for Dianna Zebib

Question

The autoionization of sulfuric acid can be described by the following chemical equation: H2SO4 (l) + H2SO4 (l) ⇌ H3SO4+(aq) + HSO4− (aq) At 25°C, K = 3 × 10−4. a. Write an equilibrium constant expression for K H2SO4 that is analogous to Kw.

Given that the density of H2SO4 is 1.8 g/cm3 at 25°C. b. What is the concentration of H3SO4+ ? c. What fraction of H2SO4 is ionized?

Answer 1

K=[H3SO4⁺]*[HSO4^-]/[H2SO4]² = 3*10^-4 ;

If the concentration of H2SO4 was 1 mole/L ,

K=[H3SO4⁺]*[HSO4^-]=3*10^-4;

From the equation emerges that [H3SO4+]=[HSO4-]; so,

[H3SO4+]=[HSO4-]=sqrt(3*10^-4)=0.0173 M

b.If the quantity of H2SO4 is 1 mole , meaning the mass is

m=n*Mr; m=98 grams;

volume will be -

V=m/ρ ; V=98/1.8=54.44 cm³=0.05444 L;

C(H3SO4+)=0.0173/0.05444=0.318;

c. mole fraction(of ionized H2SO4)=0.0173*2/1=0.0346