Question

For a weak acid, HA,

1) write the ionization reactionin water

2) write the expressio of its acid ionation constant, Ka

3)If Ka is 1.0 x 10^-6, calculate the pH of a 0.200-M Ha solution

4)What is the pH of a buffer solution with 0.35 M HA and 0.40 M A-?

Answer 1

(1) The ionization of HA in water is HA + H₂O H₃O⁺ + A^-

(2) Let a be the dissociation of the weak acid
                           HA + H₂O H₃O⁺ + A^-

initial conc.          c                 0         0

change              -ca                  +ca      +ca

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

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

                             K_a = c a² / (1-?) This is the expression for ionation constant

(3) In the case of weak acids ? is very small so 1-a is taken as 1

So K_a = ca²

a = ? ( K_a / c )

Given K_a = 1.0x10^-6

          c = concentration = 0.200 M

Plug the values we get a = 2.24x10^-3

So [ H₃O⁺ ] = ca = 0.200 x 2.24x10^-3 = 4.47x10^-4 M

pH = - log [ H₃O⁺ ]

     = - log( 4.47x10^-4 )

     = 3.35

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

          = - logK_a + log([A^-]/[HA])

         = - log(1.0x10^-6) + log(0.40/0.35)

        = 6 +0.058

       = 6.058