Question

Calculate the pH when 2.63 g of C₆H₅COONa is added to 41 mL of 0.50 M benzoic acid, C₆H₅COOH. Ignore any changes in volume. The K_a value for C₆H₅COOH is 6.5 x 10^-5

Answer 1

Molar mass of C₆H₅COONa = (7xAt.mass of C ) +(5xAt.mass of H ) + (2xAt.mass of O ) + At.mass of Na

                                             = (7x12) + ((5x1) + (2x16) + 23

                                             = 144 g/mol

Given mass of C₆H₅COONa is = 2.63 g

So Number of moles of C₆H₅COONa , n = mass/molar mass

                                                                   = 2.63 g / 144(g/mol)

                                                                   = 0.0183 mol

Number of moles of benzoic acid , n' = Molarity x volume in L

                                                           = 0.50 M x (41 /1000 ) L

                                                           = 0.0205 mol

Given K_a = 6.5x10^-5

pK_a = - log K_a

        = - log (6.5x10^-5)

        = 4.19

According to Henderson's equation ,

pH = pK_a + log ([salt] / [ acid] )

     = 4.19 + log ( 0.0183 / 0.0205 )

    = 4.14

Therefore the pH of the solution is 4.14