a 30.0ml sample of diprotic acid is titrated with .260M KOH. if 68.5ml is required to reach the equivalence point, what is the concentration of the acid?

H2A + 2KOH ==> K2A + 2H2O

moles KOH = M x L = ?
Convert moles KOH to moles H2A using the coefficients in the balanced equation. moles H2A = 1/2 mole KOH.

Then M H2A = moles H2A/L H2A. Solve for L and convert to mL.

To find the concentration of the diprotic acid, we can use the concept of stoichiometry in acid-base titration.

First, let's determine the number of moles of KOH that reacted with the diprotic acid.

The volume of KOH used is 68.5 ml, which is the same as 0.0685 L (since 1 ml = 0.001 L).

Now we can calculate the number of moles of KOH using the formula:

Moles of KOH = Volume (L) x Concentration (M)

Moles of KOH = 0.0685 L x 0.260 M
Moles of KOH = 0.01781 mol

Since the reaction between diprotic acid and KOH is 1:2 (1 mole of acid reacts with 2 moles of KOH), the number of moles of diprotic acid is twice the number of moles of KOH.

Moles of acid = 2 x Moles of KOH
Moles of acid = 2 x 0.01781 mol
Moles of acid = 0.03562 mol

Now we can calculate the concentration of the acid.

Concentration of acid (M) = Moles of acid / Volume (L)

The volume of the acid sample is given as 30.0 ml, which is the same as 0.0300 L.

Concentration of acid (M) = 0.03562 mol / 0.0300 L
Concentration of acid (M) = 1.187 M

Therefore, the concentration of the diprotic acid is 1.187 M.