To learn how to calculate ion concentrations in an aqueous solution of a strong diprotic acid.

sulfuric acid, h2so4, is a strong acid. its complete dissociation in aqueous solution is represented as

h2so4→h++hso4−

a hso4− anion can dissociate further by

hso4−⇌h++so42−

but the extent of dissociation is considerably less than 100%. the equilibrium constant for the second dissociation step is expressed as

ka2=[h+][so42−][hso4−]=0.012

part a

calculate the concentration of h+ ions in a 0.010 mol l−1 aqueous solution of sulfuric acid.

express your answer to three decimal places and include the appropriate units.

[h+] =

The concentration of H⁺ ions is 0.0165 M.

Explanation:

Let's consider the dissociation of H₂SO₄. In the first step, H₂SO₄ acts as a strong acid, completely dissociating into HSO₄⁻ and H⁺. Therefore, the concentrations of these ions will be the same that the initial concentration of the acid.

           H₂SO₄ ⇒ HSO₄⁻ +    H⁺

Initial   0.010M        0            0

Final         0        0.010M   0.010M

Now, HSO₄⁻ is a weak acid that will dissociate partially to form H⁺ and SO₄²⁻.

HSO₄⁻ ⇄ H⁺ + SO₄²⁻

To find out the concentration of H⁺ from HSO₄⁻ we will use an ICE Chart. We recognize 3 stages: Initial, Change and Equilibrium, and complete each row with the concentration or change in concentration.

    HSO₄⁻ ⇄ H⁺ + SO₄²⁻

I   0.010        0        0

C     -x          +x      +x

E  0.010 -x    x         x

This quadratic equation has 2 solutions: x₁ = -0.018 and x₂ = 0.00649. Since concentrations cannot be negative, we choose x₂. Then, [H⁺] coming from HSO₄⁻ is 0.00649 M.

The total concentration of H⁺ is:

[H⁺] = 0.010 M + 0.00649 M = 0.0165 M

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.

each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i