Acertain weak acid, ha, has a ka value of 6.7 x 10^-7.
part a calculate the percent ionization of ha in a 0.10 m solution. express your answer to two significant figures and include the appropriate units.
part b calculate the percent ionization of ha in a 0.010 m solution.
express your answer to two significant figures, and include the appropriate units.

Explanation:

Percent ionization is the percent of the original acid that has ionized:

Part A:

1) Data:

2) Equilibrium equation:

3) ICE (initial, change, equilbirium) table

                           Concentrations

                         HA            H⁺       A⁻

Initial                 0.10           0        0

Change             - x            + x     + x

Equilibrium     0.10 - x         x         x

                        6.7 × 10 ⁻⁷   =          x² / (0.10 - x)

4) Solve the equation:

Since Ka << 1, you can assume x << 0.10 and 0.10 - x ≈ 0.10

Part B:

1) Data:

2) Equilibrium equation:

3) ICE table:

                            Concentrations

                           HA              H⁺       A⁻

Initial                 0.010            0         0

Change               - x              + x     + x

Equilibrium     0.010 - x           x         x

                        6.7 × 10 ⁻⁷   =          x² / (0.010 - x)

4) Solve the equation:

Since Ka << 1, you can assume x << 0.010 and 0.010 - x ≈ 0.010

The percent ionization of HA is 2.6%.

The ICE table is as follows;

      HA(aq) ⇄ H^+(aq)     +   A^-(aq)

I      0.1            0                     0

C     -x              +x                   +x

E    0.1 - x          x                     x

If the Ka = 6.7 X 10^-7

Ka = [H^+] [A^-]/[HA]

6.7 X 10^-7= [x] [x]/[0.1 - x]

6.7 X 10^-7([0.1 - x) = x^2

6.7  X 10^-8 - 6.7 X 10^-7x =  x^2

x^2 + 6.7 X 10^-7x - 6.7  X 10^-8 = 0

x = 0.00026 M

The percent ionization is obtained from; [A^-]/[HA] X  100/1

=  0.00026 M/ 0.010 M  X  100/1 = 2.6%

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the answer is a! (2.9 x 10^-18 j) hope i could someone. : )