Question

what mass of sodium benzoate must be dissolved in 500 ml of .65 M benzoic acid to produce a buffer with the pH of 4.7

Answer 1

According to Henderson's Equation

Given pH = 4.7

pK_a of benzoic acid is = 4.20

No . of moles of benzoic acid added is , n = Molarity x Volume in L

                                                    = 0.65M x 0.5 L

                                                = 0.325 moles

Let n be the number of moles of sodium bezoate added then

Plug the values in the above Henderson's Equation we get

n = 1.028 mol

mass of sodium bezoate added , m = Number of moles x molar mass

                                                    = 1.028 mol x 144 g mol^-1

                                                    = 148.03 g

5.22 = 4.74 + log ( x / 0.96 )

                                                                          log ( x/ 0.96 ) = 0.48

                                                                                  x / 0.96 = 10^0.48

                                                                                                = 3.0199

                                                                                             x = 2.9 moles per 2.4 L

Mo .of moles ,x = mass / Molar mass

Molar mass of NH4Cl = 14 + 4 * 1 + 35.5 = 56.5 g / mol

So mass of te NH4Cl , m = x * 56.5

                                        = 2.9 * 56.5

                                        = 163.85 g of NH4Cl