Question

A 250mL buffer is made of .33 M benzoic acid and .21 M sodium benzoate. What is the initial pH of the solution? If 1.0 mL of 12M NaOH is added to the solution, what is the new pH? Then if the concentrations in part A of this experiment had been 0.50 M for the acid and 1.5 M for the base, what would the pka have been, assuming the pH remains the same?

Answer 1

Ka of Benzoic acid = 6.46*10-5

pKa= -log (6.46*10-5)=4.189

pH= pKa+ log [A-]/[HA]

A- = C6H5COO- ( from Benzoic acid ) and 0.21M   and [HA] =0.33

pH= 4.189+log(0.21/0.33)=3.99

when NaOH is added Benzoic acid beomes reacts to from Sodium Benzoate and this reduces concentration of Benzoic acid and increases the concentration of sodium Benzonate.

moles of Benzoic acid =0.33*0.25=0.0825 moles and moles of sodium benzoate= 0.21*0.25=0.0525

moles of NaOH= 12*1/1000= 0.012

moles of sodium acetate= 0.0525+0.012=0.0645 moles of benzoic acid =0.0825-0.012=0.0705

Volume after mixing = 250+1= 251ml =0.251 L

Concentrattions= sodium Benzoate= 0.0645/0.251=0.26 Benzoic acid =0.0705/0.251=0.281

pH= Pka+ log [A-]/[HA] = 4.189+log[ 0.26/0.281]=3.992=4.155

for [HA] =0.5M [A-] =1.5M

pH= pKa+log[1.5/0.5]

3.99= pKa+log3

pKa= 3.5