Question

calculate the ph of a solution when 15ml of 0.5M naoh is added to 30ml of a 0.8 M benzoic acid (Benzoic acid is a monoprotic acid and has a ka of 6.5 x 10^-5

Answer 1

Solution :-

When NaOH is added to the benzoic acid solution then it will form the benzoate ions therefore the solution forms the buffer

Lets first calculate the moles of the benzoic acid and NaOH

Moles of benzoic acid = molarity * volume in liter

                                       = 0.8 mol per L * 0.030 L

                                       = 0.024 mol

Moles of NaOH = 0.5 mol per L * 0.015 L = 0.0075 mol NaOH

So moles of benzoic acid are excess

Now lets calculate the moles of the benzoic acid remain after reaction

Moles of benzoic acid remain = 0.024 mol – 0.0075 mol = 0.0165 mol

Moles of benzoate formed = 0.0075 mol

Lets calculate the new molarities at the total volume

Total volume = 30.0 ml + 15.0 ml = 45.0 ml = 0.045 L

[benzoic acid ] = 0.0165 mol / 0.045 L = 0.367 M

[benzoate ] = 0.0075 mol / 0.045 L =0.167 M

Now using the Henderson equation we can calculate the pH

pH= pka + log ([Base]/[acid])

pka = -log ka

pka = - log 6.5*10^-5

pka = 4.187

lets put the values in the formula

pH= pka + log ([Base]/[acid])

pH= 4.187 + log ([benzoate]/[benzoic acid])

pH= 4.187 + log [0.167/0.367]

pH= 3.85

Therefore the pH of the solution is 3.85