The data below are from a titration experiment where acetic acid, HC2H3O2, was t

Question

The data below are from a titration experiment where acetic acid, HC2H3O2, was titrated with sodium hydroxide. ( The acetic acid solution are corefully pipetted into an Erlenmeyer flask, and sodium hydroxide was delivered from a burret. The equivalence point was detected by a pink phenolphthalein endpoint.)

HC2H3O2(aq) + NaOH(aq) -> NaC2H3O2(aq) + H2O(l)

Volume of acetic acid: 15.00 ml

Concentration of NaOH: 0.1456 M

Initial volume of NaOH in burret: 2.65 ml

Final Volume of NaOH in burret: 11.25ml

What is concentration of the acid?

Explanation / Answer

Volume of 0.1456 M NaOH used = 11.25 mL - 2.65 mL = 8.6 mL = 8.6 mL x (1L / 1000 mL) = 8.6x10-3 L

Hence moles of NaOH used = MxV(L) = 0.1456 mol/L x 8.6x10-3 L = 1.25216x10-3 mol NaOH

From the balanced chemical reaction it is clear that 1 mol NaOH reacts with 1 mol of HC2H3O2(aq).

Hence moles of acetic acid, HC2H3O2(aq) in the solution = (1 mol acid / 1 mol NaOH) x 1.25216x10-3 mol NaOH

= 1.25216x10-3 mol acid.

Given the volume of acid = 15.00 mL = 15.00 mL x (1L / 1000 mL) = 0.015 L

Hence concentration of the acid = moles of acid / V(L)

= 1.25216x10-3 mol acid / 0.015L = 0.0835 M (answer)

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