A multiple choice question... (1 Viewer)

[Aerlinn]

NaOCl is completely dissociated in water to form Na+(aq) and OCl-(aq). In solution OCl- hydrolyses according to the equation
OCl-(aq) + H20(l) <---> HOCl(aq) + OH-(aq)
100ml of pure water at constant temperature is added to a 100ml solution of 0.10M NaOCl. When the solution reaches equilibrium,
A. the H+ concentration has decreased
B. the pH of the solution has decreased
C. the concentration of HOCl has increased
D. the value of the equilibrium constant has halved.
(answer: B. I'm confused as to why that's right, other other options, ie. C are wrong...)

 

[xiao1985]

adding water has negligible effect on the equilibrium and conc. of aq substances...

which is why, although conc of HOCl does increase, the magnitude is rather small...

this is because since the reaction is done in water in the first place, [H2O] = 55 molar (yes, i did the calculation), and adding more water would not increase the concentration much, hence equilibrium is not disturbed...

on the other hand, adding more solvent does change [OH-] and consequently influences pH

 

[Aerlinn]

Adding water decreases [OH-] I suppose, but how does that affect [H+]?...

For OCl-(aq) + H20(l) <---> HOCl(aq) + OH-(aq) and picking C, maybe I was thinking of acids and ionisation, that when you add water, it ionises more? Or maybe I'm completely wrong. I don't know...

 

[xiao1985]

errr, in normal conditions, pH +pOH = 14... [OH] gives pOH, and you can figure out pH that way...

well, it won't ionises more... i am thinking you are trying to logic it out using le chatelier eh? well in this question, le chatelier doesn't work in the way you wanted it... conc of water is more or less constant