Ka and Kb in Chem.?

Question

I have no idea what Ka or Kb even is. I just need some help.

1. When 0.20 mol of lactic acid, HC3H5O3. is added to water to make 1.0 L of solution. The equilibrium hydrogen ion concentration is 5.3 x 10-3 M. What is the Ka of the lactic acid?

HC3H5O3 (aq) → H+(aq) + C3H5O3- (aq)

2. A 0.10M solution of formic acid has an equilibrium [H+] = 4.2 x 10-3 M. What is the Ka of the formic acid?

HCOOH(aq) → H+(aq) + HCOO- (aq)

3. What is the equilibrium hydrogen ion concentration of a 0.25M solution of carbonic acid, H2CO3? The Ka = 4.2 x 10-7.

H2CO3 (aq) → H+(aq) + HCO3-(aq)

4. What is the equilibrium for the hydroxide ion in a 0.10M solution of CN-? The Kb is 2.0 x 10-5.

CN-(aq) → OH-(aq) + HCN(aq)

5. What is the equilibrium concentration of hydrogen ion in a 0.10 M solution of propanoic acid, C2H5COOH? The Ka of propanoic acid is 1.8 X 10-5.

C2H5COOH(aq) → H+(aq) + C2H5COO -(aq)

Any help is appreciated

Answer 1

Where have you been? Sleep through the last lecture?

They are equilibrium constants for weak acids (Ka) and (Kb). Such acids, when dissolved in water, ionize only slightly. For that case, the extent of ionization is given by Ka or Kb. For the weak monobasic acid HA, at equilibrium,
[H+][A-]/[HA]= Ka.
MOST OF THE TIME, in your calcs [HA] will be approximately the initial HA conc.

Try prob 2. From the equation above, the formate ion conc also equals 4.2x10-3M. This is sufficiently small compared to 0.1 M formic acid, that you can use 0.1 in the calc. (purists would use 0.0952). So the calc for the Ka is
4.2x10-3 * 4.2x 10-3 / 0.1 = 1.7x10-4
In problem 4, the CYANIDE ION is the base. When the equation is in this form
[OH-][HCN]/ [CN-] = Ka
Here the initial [CN-] is 0.1M, and only a small amount of this reacts to reach equilibrium. Call that "x". From the equation,
[x] [x]/0.1 = Kb = 2x10-5
x= 1.4x10-3. Again this is "small" compared to 0.1.

Answer 2

that's an interesting question

Answer 3

It depends..