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Acids and Bases - Calculations ...

Difficult difficult question ....

I'm stuck on the second part to a question on calculating a volume of substance needed for equivalence in a titration ... just don't see that i have enough information to solve it, and where to start/what to do ... any help would be much appreciated ... thankyou in advance.

Please refer to attachment xx

Acid Base Calculations.doc20.0KB

Reply 1

charco

Study Forum Helper

The pH of solution A, a 0.15 mol dm-3 solution of a weak monoprotic acid HX, is 2.69.
Calculate [H+] in solution A and hence determine the value of the acid dissociation constant, Ka, of HX.

I got a [H+] value of 2.04 * 10-3

I got a Ka value of 2.78 * 10-5 mol dm-3

A 25cm3 sample of A is titrated with 0.25 mol dm-3 sodium hydroxide. Calculate the volume of sodium hydroxide needed to reach equivalence in the titration. Give the best estimate you can of the pH of the neutralised solution, stating a reason.

I really don’t see how to begin this question. I’d be tempted to use the equation for number of moles = MV / 1000, but I can’t see enough information for this.

I also don’t see how I would give an estimate of the neutralised solution, or the reason – why won’t it be a certain figure?

Ka = [H+][A-]/[HA]

as its a weak acid assume that [HA] is the same as the initial

Ka = [H+]^2 /0.15
pH = 2.69 therefore [H+] = 0.00204
Ka = 2.73 x 10^-5

for the titration you are told that its a monoprotic acid - it's a 1:1 moles relationship.

0.15 x 25 = 0.25 x volume of the NaOH
volume of the NaOH = 15cm3

to estimate the pH you now only have sodium X solution.

By hydrolysis this will be basic i.e. pH greater than 7 because it is the salt of a strong base and a weak acid. It is difficult to arrive at an exact figure as there are two separate equilibria set up by the hydrolysis.

NaX <==> Na+ + X-
H2O + X- <==> HX + OH-

the first equilibrium lies almost 100% to the right hand side (its an ionic compound)
the second depends on the value of Ka for the weak acid (2.69)

If you wished to calculate an approx pH you would have to apply the acid equilibrium equation to this second equilibrium calculating values for all the components by considering the number of moles an the total volume of the mixture.

i.e the number of moles of NaX will be 0.15 x 0.025
the total volume is 0.040 litres
so the concentration is 0.09375

assuming that all of the NaX dissociates this would initially give [X-] of the same...
but this almost 100% turns into HX so you would have to assume that this is the [HX] value

now use Ka = [H+][X-]/[HX]

etc etc

so with a few approximations you get to a pH value