Question

i have 2 chemistry questions. please give me correct answers. thank you .

1. Calculate the pH of a 0.23 M solution of acetic acid (CH3COOH).

DATA : pKa (CH3COOH) = 4.76

Answer:

2.

Calculate the pH of a 0.37 M solution of potassium cyanide (KCN).

DATA : pKa (HCN) = 9.21

Answer:

Answer 1

(1)

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

initial conc.            c               0         0

change                -ca            +ca      +ca

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 a is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given pK_a = 4.76

   - log K_a = 4.76

         K_a = 10^-pKa

         K_a = 1.74x10^-5

          c = concentration = 0.23 M

Plug the values we get a = 8.69x10^-3

[H⁺] = ca = 0.23x8.69x10^-3 = 2.00x10^-3 M

pH = - log[H⁺] = - log( 2.00x10^-3)

     = 2.7

Simillarly the second one

(1)

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

initial conc.            c               0         0

change                -ca            +ca      +ca

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 a is very small so 1-a is taken as 1

So K_a = ca²

==> a = √ ( K_a / c )

Given pK_a = 9.21

   - log K_a = 9.21

         K_a = 10^-pKa

         K_a = 6.16x10^-10

          c = concentration = 0.37 M

Plug the values we get a = 4.08x10^-5

[H⁺] = ca = 0.37x4.08x10^-5 = 1.51x10^-5 M

pH = - log[H⁺] = - log( 1.51x10^-5)

     = 4.82