A2 Chemistry: Acid Dissociation Constant

Solution A contains n moles of a different weak acid, HX. The addition of some sodium hydroxide to A neutralises one third of the HX present to produce solution B.

i) In terms of the amount, n, how many moles of HX are present in solution B [1]

I think the answer is n - 1/3n

ii) Determine the ratio [HX]/[X-] in Solution B. [2]

I am very uncertain but a wild guess would be [n]/ [2/3n]. Is that correct?

iii) Solution B has a hydrogen ion concentration of 4.2 x 10^4 mol dm^-3. Use this information to determine the value of the acid dissociation constant of HX. [2]

I am stuck could someone please show me?

PS: Sorry for being a pain.

Reply 1

oxymoron

i) yes n-1/3n = 2/3n

ii) from i), [HX]=2/3n ... and [H-] = 1/3n (from neutralised acid)

therefore [HX]/[X-] = (2/3n)/(1/3n) = 2/1 = 2

iii) Ka=[H+][X-]/[HX] but we know [X-]/[HX]=1/2 from ii)

so Ka = [H+] * 1/2 = 2.1 x 10^4 (are you sure that isn't a minus 4 by the way ... that doesn't sound like a weak acid!)