Question

We dissolve 0.500 g of an organic acid `C_6H_10O_4` in water and titrate the solution with 0.250 mol/L KOH. It takes 27.4 mL of KOH to neutralize the organic acid. It is a monoprotic or polytropic acid?

Answer 1

It is a diprotic acid.

Explanation:

Step 1. Calculate the moles of `"KOH"`

`"Moles of KOH" = 0.0274 color(red)(cancel(color(black)("L KOH"))) × "0.250 mol KOH"/(1 color(red)(cancel(color(black)("L KOH")))) = "0.006 850 mol KOH"`

Step 2. Calculate the moles of acid

`"Moles of acid" = 0.500 color(red)(cancel(color(black)("g C"_6"H"_10"O"_4))) × ("1 mol C"_6"H"_10"O"_4)/(146.14 color(red)(cancel(color(black)("g C"_6"H"_10"O"_4)))) = "0.003 421 mol C"_6"H"_10"O"_4`

Step 3. Calculate the molar ratio of base to acid

`"Molar ratio" = ("0.006 850" color(red)(cancel(color(black)("mol"))))/("0.003 421" color(red)(cancel(color(black)("mol")))) = 2.002 ≈ 2`

It takes 2 mol of the base to neutralize 1 mol of acid, so the acid must be diprotic.

The equation for the reaction is

`"H"_2"C"_6"H"_8"O"_4 + "2KOH" → "K"_2"C"_6"H"_8"O"_4 + 2"H"_2"O"`