Question

A certain weak acid, HA, has a Ka value of 8.4�10^?7.

Part A

Calculate the percent ionization of HA in a 0.10 M solution.

Part B

Calculate the percent ionization of HA in a 0.010 M solution.

Answer 1

The ionization equation of acid HA is as follows: HA → H+ + A The percent ionization is given; at equilibrium x 100% [HA] tatal % ionization-

Part A: The acid initial concentration (HAis 0.10 M. Set ICE table to find the equilibrium concentrations Initial(M) Change (M) Equilibrium(M): 0.10-xx 0.10 Substitute the equilibrium values in acid ionization constant expression κ = [A][r] НА 8.4x10-7 (x)(x) Since 8.4x10- <<0.10, assume that 0.10-x * 0.10 8.4×10-7 x = 2.9 x 10-4 so, IH' | = 2.9 x 10-4 M % ionization = 2.SK 0.10-x (0.10) ×100% 0.29% = 0.10

Part B The acid initial concentration (HAis 0.010 M. Set ICE table to find the equilibrium concentrations HA-H + A Initial(M) Change (M) Equilibium(M): 0.010-x x Substitute the equilibrium values in acid ionization constant expression 0.010 צל צל AH K,=--[HA] rar) Since 8.4x10-. << 0.010. assume that 0.010-x ~ 0.010 4x10 T0.010) x9.2x10-5 So, [H.] = 92x10-5 M % ionization ×100% 0.92% = 0.010

Answer 2

Dissociation equation of the acid;

HA(aq) <-----> H+(aq) + A-(aq)
0.010-x M ........ x M ........ x M

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

1.3x10^-7 = (x)(x) / (0.010 - x)

Since the amount of x is very small compared to 0.010 M, it is neglected.
1.3x10^-7 = (x)(x) / (0.010)
x^2 = 1.3x10^-9
x = 3.6x10^-5 M

Since 3.6x10^-5 M is the amount dissociared (ionized)
and 0.010 M is the initial amount,
the percent ionization of the acid will be:
(3.6x10^-5 M / 0.010 M) x 100 = 0.36 %

If this amount is greater than 5%, we cannot apply the approximation of neglecting (x). In such cases, the roots of the quadratic equation must be found. (One of the roots will satisfy the condition).

Answer 3

percent dissociation = amount of acid ionized / total initial concentration of acid x 100.

HA --> H+ + A-
.10 0.0 0.0 I
-x +x +x C
.10-x x x E

I don't know if you know this table, but our book calls it the ICE method (Initial-Concentration-Equilibrium). The first line is the initial concentration of everything before the acid begins to ionize (form H+). This is 0.10 M--the amount of acid. Now H+ isnt necessarily 0.0, however, we are approximating (remember the acid is in water, which self ionizes--technically theres 1.0 x 10^14 H+ ions initally, but that number is so small, we can ignore it.). p.s. H30+ and H+ are used interchangeably. Now we want to see the change, since everything has a coefficient of 1, its all the same concentrations being added and subtracted. Now we create an equilibrium equation.

Ka = [H+][A-]/[HA]
1.3*10-7 = (x)(x) / (.10 -x)
Since, 1.3*10^-7 is much much smaller in comparison to .10, you can neglect the x its subtracted from to 0. Therefore.
1.3*10^-7 = x^2 / .10

Now you can solve for x. Once you have x, this is the amount of acid ionized. Divide this number by the initial concentration of acid (.10) and times by 100. This is your percent dissociation. It should be relatively low