Question

In a titration, 2.7 cm3 of 0.100 mol dm-3 sodium hydroxide, NaOH, solution is added to 25.0
cm3 of 0.125 mol dm-3 benzoic acid, C6H5COOH, solution. Calculate the pH of the
resulting solution given that the pKa of benzoic acid is 4.19.​

Answer 1

Create a reaction between NaOH and H2SO4. And then, complete the coefisien reaction below.

Reaction :

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

Step 2. Convert mol/dm^3 to mmol/cm^3

(0.25 mol x 1000 mmol/mol) / (dm^3 x 1000 cm^3/dm^3)

0.25 mmol/cm^3

Step 3. Find out mmol of NaOH.

M : mmol/cm^3

mmol : M x cm^3

mmol of NaOH : 0.25 mmol/cm^3 x 25 cm^3

NaOH : 6.25 mmol

Step 4. By using data of coefisien reaction. We are going to find out the mmol of H2SO4

2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)

mmol of H2SO4 equals mmol of NaOH

mmol of H2SO4 : (Coefisien of H2SO4/Coefisien of NaOH) x mmol of NaOH

mmol of H2SO4 : (1 / 2) x 6.25 mmol

H2SO4 : 3.125 mmol

Step 5. Finally, we are going to find out the concentration of H2SO4.

M : mmol / cm^3

M of H2SO4 : 3.125 mmol / 25 cm^3

H2SO4 : 0.125 mmol/cm^3 or 0.125 M

So, the concentration of H2SO4 is 0.125 M

hope this will help you..

-3idiots29