H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second proton is 1.1x10-2. What would be the pH of a solution that is 0.100 M H2SO4? Account for the ionization of both protons.

Question options:

a) 2.05

b) 0.963

c) 0.955

d) 1.00

b) 0.963

Explanation:

H2SO4→ HSO4- + H3O+

HSO4- + H2O ⇌ SO42- + H3O+

Construct ICE table:

        HSO4- (aq)    +    H2O        ⇌      SO42- (aq)     +     H3O+ (aq)

I          0.1                  solid &                   0                          0.1

C         -x                     liquid                 + x                            + x

E         0.1 - x          are ignored              x                          0.1 + x

Calculate x

Ka = products/reactants

  =

0.011 =

0.011 x (0.1 -x) = o.1x + x^2

0.0011 - 0.011 x - o.1x - x^2 = 0

0.0011 - 0.011 x - x^2 = 0

Use formula to solve for quadratic equation

x = / 2a

a = -1, b = -0.111, c = 0.001

Solve for x

x =  / 2(-1)

x = 0.111 +,- / -2

x = 0.111 +,- / -2

x =

x =   , x =

x =    , x =

x = - 0.12015  , x = 0.00915

x cannot be negative, so

x = 0.00915 M

Calculate [H3O+]

[H3O+] = 0.1 M + x

[H3O+] = 0.1 M + 0.00915 M

[H3O+] = 0.10915 M

Clculate pH

pH = - log [ H3O+]

pH = - log [ 0.10915]

pH = 0.963

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.