A chemist titrates 220.0 mL of a 0.1917M propionic acid (HC2H5CO2) solution with 0.1787 M KOH solution at 25°C. Calculate the pH at equivalence. The pKa of propionic acid is 4.89.
Round your answer to 2 decimal places.

Answer : The pH at equivalence is, 9.08

Explanation : Given,

Concentration of = 0.1917 M

Volume of = 220.0 mL = 0.220 L (1 L = 1000 mL)

First we have to calculate the moles of

As we known that at equivalent point, the moles of and KOH are equal.

So, Moles of KOH = Moles of = 0.0422 mol

Now we have to calculate the volume of KOH.

Total volume of solution = 0.220 L + 0.00754 L = 0.22754 L

Now we have to calculate the concentration of KCN.

The balanced equilibrium reaction will be:

Moles of = 0.0422 mol

At equivalent point,

Given:

Now put all the given values in the above expression, we get:

Therefore, the pH at equivalence is, 9.08

we need the questions , i may then you.