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Titration calculation : Quick Check

Hi,

Could someone check this for me?

A titre of 23.45cm³ was obtained when ethanoic acid was titrated against 25.00cm³ of 0.100moldm^-3 sodium hydroxide. Find the concentration of the ethanoic acid in mol/dm³ and g/dm³.

n(NaOH) = c x v

= 0.100 x (25 x 10^-3)

= 0.0025 moles therefore 0.0025 moles ethanoic acid.

conc(CH₃COOH) = n/v

= 0.0025/0.002345

= 0.107moldm^-3

conc in gdm^-3 = conc in moldm^-3 x Mr

= 0.107 x 60.0

= 6.40gdm^-3

Thanks!

Reply 1

_anum

looks right to me :smile:

Reply 2

nexttime

well from the initial volumes given you'd expect the answer to be slightly more concentrated than the NaOH if its a 1:1 reaction, so would seem about right :dontknow:

Reply 3

CHEM1STRY

nexttime

well from the initial volumes given you'd expect the answer to be slightly more concentrated than the NaOH if its a 1:1 reaction, so would seem about right :dontknow: