Question

If 4.1 g of butanoic acid, C4H8O2, is dissolved in enough water to make 1.0 L of solution, what is the resulting pH?

Answer 1

Molar mass of butanoic acid, C4H8O2 = (4x12) + (8x1) + (2x16) = 88 g/mol

So number of moles , n = mass/molar mass

                                   = 4.1 g / 88 (g/mol)

                                   = 0.0466 mol

Molarity of butanoic acid , M = number of moles / volume in L

                                         = 0.0466 mol / 1.0 L

                                         = 0.0466 M

Let a be the dissociation of the weak acid,butanoic acid
                            HA <---> H ⁺ + A^-

initial conc.            c               0         0

Equb. conc.         c(1-a)          ca       ca

Dissociation constant , K_a = ca x ca / ( c(1-a)

                                         = c a² / (1-a)

In the case of weak acids α is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given K_a = 1.5x10^-5

          c = concentration = 0.0466 M

Plug the values we get a = 0.0179

So [H⁺] = ca = 0.0466 x 0.0179 = 8.36x10^-4 M

pH = - log [H⁺]

     = - log (8.36x10^-4 )

     = 3.08

Therefore the pH of the solution is 3.08