Questions i had the most trouble with (1 Viewer)

[lanvins]

1. The OCl- ion acts as a base in water according to the equation
OCl-(aq) + H2O(l) HOCl(aq) + OH-(aq)
When two drops of 5.0 M NaOH are added to an equilibrium mixture of OCl in water at constant
temperature what is the effect on equilibrium?

2. Benzoic acid, C6H5COOH, molar mass 122 g mol-1, is a weak monoprotic acid.
The pH of a solution formed when 500 mg of benzoic acid is dissolved completely in water to form 200 mL
of solution is?

3. Phosphorus reacts with excess chlorine according to the following equation.
P4(s) + 10Cl2(g) → 4PCl5(g)
When 6.49 g of PCl5(g) is produced, 11.7 kJ of energy is released.
ΔH for this reaction, in kJ mol-1, is?

4. The equation for the self ionisation of water is given below.
2H2O(l) H3O+(aq) + OH-(aq) ΔH = + 57 kJ mol-1
At 90°C, water is?

thanks

 

[brenton1987]

1. Equilibrium is pushed to the left to produce OCl and H₂O.
2. pH is < 7. A quantitative answer cannot be given based on the information.
4. Hot? The question isnt really asking anything. Ionisation will increase because the reaction is endothermic and heat has been added.

There is too many calculations in 3 for me to do it right now. Ill post an answer for it soon.

 

[lanvins]

thanks

 

[JMill]

For Question 2, how can u calculate the pH if you don' tknow the equalibrium constant? it is impossible to calculate the pH without knowing the amount of mols oh H+ ions in solution and without the K value the concentration of H+ cannot be calculated as the acid is weak (not completely ionised). `:angry:

 

[rufous]

Regarding water at 90 degrees, it will be neutral. Even though dissociation is increased, the concs, of H+ and OH- are still equal!