Question

Calculate the values of pH when 40.0 mL of 0.0250M benzoic acid (HC7H5O2 , Ka=6.3x10-5 ) is titrated with (1) 0.0 mL of 0.050 M NaOH solution? (2) 10.0 mL of 0.050 M NaOH solution? (3) 25.0 mL of 0.050 M NaOH solution?

I mostly need help with 2 and 3. The answers are 4.20 and 11.58 but I not know how to calculate them.

Answer 1

moles of NaOH added= molarity* Volume in L=0.05*10/1000 =0.0005

moles of Benzoic acid ( designated as HA). = 0.0250*40/1000 =0.001

the reaction between Benzoic acid and NaOH is C6H5COOH +NaOH ------->C6H5COONa+ H2O

1 mole of NaOH reacts with 1mole of Benzoic acid. Molar ratio of C6H5COOH: NaOH= 1:1 ( theoretical)

Actual molar ratio =0.0005:0.001 =1:2

excess is Benzoic acid . All the NaOH gets consumed and moles of sodium Benzoate ( C6H5COONa) formed =0.0005, moles of Benzoic acid remaining =0.001-0.0005=0.0005

Volume of solution after mixing = 40+10= 50ml= 50/1000L= 0.05L

concentrations: Benzoic acid = sodium Benzonate= 0.005/0.05=0.1 M

since pH= pKa+ log [ sodium Benzoate/ Benzoic acid]

pKa of benzoic acid =-log (Ka)=4.2

hence pH= 4.2+ log (0.1/0.1)= 4.2

3. when 25ml of 0.05M NaOH is added. moles of NaOH=0.05*25/1000 =0.00125, moles of Benzoic acid =0.001

hence excess is NaOH all the Benzoic acid is consumed. Moles of NaOH remaining =0.00125-0.001= 0.00025

volume of solution after mixing=40+25= 65ml= 65/1000L=0.065L

concentration of NaOH= 0.00025/0.065=0.003846

NaOH being strong base ionizes completely. hence NaOH------>Na++ OH-

[OH-]=0.003845, pOH= 2.41, pH= 14-2.41= 11.59

pOH= -log (0.008)= 2.1