Question

Here are some problems and their answers. I would appreciate if someone could please explain step by step how to get these answers? I need to understand this concept.

The following questions refer to the equilibria: 2 NaHCO3(s) ⇌ Na2CO3(s) + H2O(g) + CO2(g) K(110°C)=0.68

1)If CO2(g) is added to the equilibrium mixture the partial pressure of H2O(g) will: answer: Decrease (how and why??)

2) If the volume of the container is decreased, the amount in grams of sodium carbonate (Na2CO3(s)) will: Answer: decrease (I thought volume only affects gases??)

3)The ΔH of the reaction was measured and determined to be -25 kJ/mole of sodium bicarbonate (NaHCO3). If the reaction were run at 150°C, a possible value for K could be: answer: 0.54

------------------------------------------------------------

This next question is not involved with the above.

4) What is the pH of the solution after 30 mL of 0.12 M HCl is mixed with 60 mL of 0.12M NaOH? Answer: . 12.60

Answer 1

Answer – Given, reaction -

1. NaHCO₃(s) <---> Na₂CO₃(s) + H₂O(g) + CO₂(g) K(110°C) = 0.68

1)If CO₂(g) is added to the equilibrium mixture the partial pressure of H2O(g) will decrease, because when we added CO₂ gas the reaction gets reversed as per Le Chatelier's principle and there is Na₂CO₃(s) ,CO₂ and H₂O gets reacted and form the NaHCO₃(s). So as the H₂O(g) reacted it moles get decrease and as mole decrease partial pressure of H₂O(g) also decreased.

2) If the volume of the container is decreased, the amount in grams of sodium carbonate (Na2CO3(s)) will decrease, because we know as the volume decrease there is mole also decrease since volume and moles are directly proportional to each other.

Answer – Given, volume = 30 mL , [HCl] = 0.12 M , volume of NaOH = 60 mL , [NaOH] = 0.12 M

We need to calculate the moles of each

Moles of HCl = 0.12 M * 0.030 L

                       = 0.0036 moles

Moles of HCl = 0.12 M * 0.060 L

                       = 0.0072 moles

Moles of HCl is limiting reactant, so excess of the NaOH

Moles of NaOH = 0.0072 – 0.0036

                         = 0.0036 moles

Total volume = 30+ 60 = 90 mL

Moles of NaOH = moles of OH^- = 0.0036 moles

[OH^-] = 0.0036 moles / 0.090 L

           = 0.040 M

So, pOH = -log [OH^-]

              = -log 0.040 M

              = 1.40

pH = 14 –pOH

      = 14 – 1.40

       = 12.60