Question

1) A 150.0 mL solution was prepared by mixing 100.0 mL of 1.00 M HONO (Ka = 4.5 x 10–4 ) with 50.0 mL of 1.00 M NaNO2 . Calculate the pH, [HONO], and [NO2–], once equilibrium has been established.

2). What is the pH, [OH^–], [HF], [F^–], and the percent ionization of a solution prepared as 0.50 M HF ? (HF: K_a = 7.2 x 10^-4)

50 M HF ? (HF: K_a = 7.2 x 10^-4)

3. Consider the reaction:

CO(g) + H₂O(g)    CO₂(g) + H₂(g)

At 800 K the equilibrium constant, K = 16.

If your reaction vessel initially contains 0.050 M CO, 0.050 M H₂O, 0.30 M CO₂ and 0.30 M H₂. What are the equilibrium concentrations of all of the components?

4. What is the pH of a saturated solution of Mn(OH)₂ ( K_sp = 2.0 x 10^–13)?

Mn(OH)₂ (s)   Mn²⁺ (aq) + 2 OH^-(aq)     

5) The weakest strong acids is Nitric acid, with a K_a = 240. The % ionizations of strong acids are rounded off to 100%. Calculate the % ionization of 0.100 M nitric acid without rounding.

Answer 1

I have a little doubt with question 1. So I'll answer question 2.

2. HF ---------> H⁺ + F^-

i) 0.5 0 0

eq) 0.5-x x x

Ka = x² / 0.5-x ----> Ka is low so 0.5-x= 0.5

x = (7.2x10^-4 * 0.5)^1/2 = 0.0189 M

The concentration of H⁺ is 0.0189 M, same with the F^-. Now, the pH

pH = -log(0.0189) = 1.72

[OH] = 1x10^-14 / 0.0189 = 5.29x10^-13 M.

PErcent of ionization is:

% = 0.0189 / 0.5 * 100 = 3.78%

3. CO(g) + H2O(g) --------> CO2(g) + H2(g) K = [H2] [CO2] / [CO] [H2O]

i) 0.05 0.05 0.3 0.3

eq) (0.05-x)² (0.3+x)²

16 = (0.3+x)² / (0.05-x)²

16 (0.0025 - 0.1x + x²) = (0.09 + 0.6x + x²)

0.04 - 1.6x + 16x² = 0.09 + 0.6x + x²

15x² - 2.2x - 0.05 = 0

x = 2.2 (2.2² + 4*15*0.05)^1/2 / 2*15

x = 2.2 2.8 / 30

x₁ = 0.17 M

x₂ = -0.02 M

The concentration of 0.17 can't be because at the end, co and h20 would be negative, so, it has to be the other value:

[CO] = 0.05 - (-0.02) = 0.07 M = [H2O]

[CO2] = [H2] = 0.3 - 0.02 = 0.28 M

4. Mn(OH)_2(s) -----------> Mn²⁺ + 2OH^-

Ksp = (s) (2s)²

Ksp = 4s³

(2x10^-13 / 4)^1/3 = s

s = 3.68x10^-5 M = [OH^-]

pOH = -log(3.68x10^-5) = 4.43

pH = 14-4.43 = 9.57

Question 5 and 1 post them per separate in a new question, so I could answer them better and faster :). Hope this helps.