Question

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 1

Answer : The pH at equivalence is, 9.08

Explanation : Given,

Concentration of
`HC_2H_5CO_2` = 0.1917 M

Volume of
`HC_2H_5CO_2` = 220.0 mL = 0.220 L (1 L = 1000 mL)

First we have to calculate the moles of
`HC_2H_5CO_2`

`\text{Moles of }HC_2H_5CO_2=\text{Concentration of }HC_2H_5CO_2* \text{Volume of }HC_2H_5CO_2`

`\text{Moles of }HC_2H_5CO_2=0.1917M* 0.220L=0.0422`

As we known that at equivalent point, the moles of
`HC_2H_5CO_2` and KOH are equal.

So, Moles of KOH = Moles of
`HC_2H_5CO_2` = 0.0422 mol

Now we have to calculate the volume of KOH.

`\text{Volume of }KOH=\frac{\text{Moles of }KOH}{\text{Concentration of }KOH}`

`\text{Volume of }KOH=(0.0422mol)/(0.1787M)`

`\text{Volume of }KOH=0.00754`

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:

`HC_2H_5CO_2+KOH\rightleftharpoons C_2H_5CO_2K+H_2O`

Moles of
`C_2H_5CO_2K` = 0.0422 mol

`\text{Concentration of }C_2H_5CO_2K=(0.0422mol)/(0.22754L)=0.1855M`

At equivalent point,

`pH=(1)/(2)[pK_w+pK_a+\log C]`

Given:

`pK_w=14\\\\pK_a=4.89\\\\C=0.1855M`

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

`pH=(1)/(2)[14+4.89+\log (0.1855)]`

`pH=9.08`

Therefore, the pH at equivalence is, 9.08