The Ka of a monoprotic weak acid is 3.39 × 10-3. What is the percent ionization of a 0.125 M solution of this acid?
I just do not know where to even start...
..........HA --> H^+ + A^-
Initial.0.125....0......0
change....-x.....x......x
equil..0.125-x...x......x
Ka = (H^+)(A^-)/(HA)
Plug into the Ka expression and solve for x = (H^+).
Then % ion = [((H^+)/(0.125)]*100 = ?
Oh, ok! I was over thinking the problem!!! Plus a fever may not help with my problem solving abilities... Thank you!
To find the percent ionization of a weak acid, you need to use the Ka value and the initial concentration of the acid. Here's how you can solve the problem step by step:
Step 1: Write the equation representing the ionization of the weak acid.
HA (aq) ⇌ H+ (aq) + A- (aq)
The acid is represented by HA, and H+ and A- represent the hydrogen ion and the conjugate base, respectively.
Step 2: Set up an ICE (Initial, Change, Equilibrium) table.
In this case, the initial concentration of HA is given as 0.125 M, so you would enter this value under the column labeled "Initial" for HA. Since HA ionizes into H+ and A-, their initial concentrations would be zero.
| HA (aq) | H+ (aq) | A- (aq)
---------------------------------------------------
Initial concentration | 0.125 | 0 | 0
Step 3: Define the equilibrium concentrations.
Let's assume that x represents the concentration of H+ and A- ions that are formed when the acid ionizes. Since one molecule of HA produces one H+ and one A-, the equilibrium concentration of HA would be 0.125 - x, while the equilibrium concentrations of H+ and A- would both be x.
| HA (aq) | H+ (aq) | A- (aq)
------------------------------------------------------
Initial concentration | 0.125 | 0 | 0
------------------------------------------------------
Change | -x | +x | +x
------------------------------------------------------
Equilibrium |0.125 - x| x | x
Step 4: Write the expression for the Ka.
The Ka expression for this weak acid can be written as:
Ka = [H+][A-] / [HA]
Since the value of Ka is given as 3.39 × 10^-3, you can substitute the equilibrium concentrations from the ICE table into the expression:
3.39 × 10^-3 = x * x / (0.125 - x)
Step 5: Solve for x.
Rearrange the equation to solve for x:
3.39 × 10^-3 * (0.125 - x) = x^2
Simplifying the equation gives:
4.2375 × 10^-4 - 3.39 × 10^-3x = x^2
Rearrange the equation to set it equal to zero:
x^2 + 3.39 × 10^-3x - 4.2375 × 10^-4 = 0
Use the quadratic formula to solve for x:
x = [-b ± √(b^2 - 4ac)] / 2a
In this case, a = 1, b = 3.39 × 10^-3, and c = -4.2375 × 10^-4. Plug in these values and solve for x. You should find two possible values for x, but you can ignore the negative value since the concentration cannot be negative.
Step 6: Calculate the percent ionization.
The percent ionization can be calculated using the equation:
%(ionization) = (concentration of H+ at equilibrium / initial concentration of acid) * 100
Plug in the value of x (the concentration of H+ at equilibrium) and the initial concentration of the acid (0.125 M) to calculate the percent ionization.
That's it! By following these steps, you should be able to find the percent ionization of the weak acid.