Question

What mass of sodium benzoate should be added to 160.0 mL of a 0.13 M benzoic acid solution in order to obtain a buffer with a pH of 4.25?

Answer 1

Data given : Volume of benzoic acid solution = 160 ml

                  Molarity of benzoic acid solution = 0.13 M

                pH= 4.25

Ka for benzoic acid =6.5*10^-3

Let benzoic acid be =[HA] and Benzoate ion be [A^-] for simplicity,

we can write the reaction as ,

[HA] ------->   [H⁺] + [A^-]

The Ka expression for the above reaction can be written as,

we have the value of Ka =6.5*10^-5 , also we know [HA] =0.13 ,

we will convert pH value to [H⁺]

pH = - log([H+])

[H+] = 10^-pH

[H+] = 10^-4.25 =5.623*10^-5 M

Substituting all the known value to obatain the concentration of [A^-]

From above equation [A^-] =0.15027 M

number of moles of sodium benzoate = 0.15027 * 160 *10^-3 = 0.024 moles

Molar mass of sodium benzoate = 144 gram/ moles

mass of sodium benzoate = moles of sodium benzoate * molar mass of sodium benzoate = 0.024 * 144 =3.4624 grams

3.46 grams of sodium should be added to 160 ml of a 0.13 M benzoic acid solution in order to obtain a buffer of pH of 4.25