I have no clue, please help and explain! =)
Find the pH of a 0.150 M solution of a weak monoprotic acid having Ka=1.2×10−5.
Find the percent dissociation of this solution.
Find the pH of a 0.150 M solution of a weak monoprotic acid having Ka=1.4×10−3.
Find the percent dissociation of this solution.
Find the pH of a 0.150 solution of a weak monoprotic acid having Ka= 0.11.
Find the percent dissociation of this solution.
I need help =(
and thank you!
To find the pH and percent dissociation of a weak monoprotic acid solution, you can follow these steps:
1. Determine the initial concentration of the acid solution. In this case, the concentration is given as 0.150 M.
2. Use the equilibrium constant (Ka) provided in the question to set up the equation for the dissociation of the acid. The generic equation for a weak monoprotic acid can be written as follows:
HA (aq) ⇌ H+ (aq) + A- (aq)
3. Write the expression for the Ka of the acid, which would be [H+][A-]/[HA]. Since we are given the value of Ka, we can assume that the concentration of H+ is equal to the concentration of A- at equilibrium.
4. Since the initial concentration of HA is given, you can assume that the concentration of A- at equilibrium would also be equal to x (as it is a monoprotic acid). Therefore, the concentration of H+ at equilibrium would also be x.
5. Substitute the known values into the Ka expression and solve for x. It would be convenient to use an ICE (Initial, Change, Equilibrium) table for this purpose.
6. Once you have determined the value of x, you can calculate the concentration of H+ by subtracting it from the initial concentration of HA. Then, convert this concentration to pH using the equation: pH = -log[H+].
7. To find the percent dissociation of the acid, use the equation: percent dissociation = (concentration of H+ / initial concentration of HA) * 100.
Now, let's apply these steps to the given scenarios:
Scenario 1:
Ka = 1.2×10^(-5)
Initial concentration of acid (HA) = 0.150 M
By solving the equations as mentioned earlier, you would find that x = 0.0109 M.
Thus, the concentration of H+ = 0.0109 M.
Using the equation pH = -log[H+], we find that pH ≈ 1.96.
To find the percent dissociation, percent dissociation = (0.0109 M/0.150 M) * 100 = 7.27%.
Scenario 2:
Ka = 1.4×10^(-3)
Initial concentration of acid (HA) = 0.150 M
By solving the equations as mentioned earlier, you would find that x = 0.033 M.
Thus, the concentration of H+ = 0.033 M.
Using the equation pH = -log[H+], we find that pH ≈ 1.48.
To find the percent dissociation, percent dissociation = (0.033 M/0.150 M) * 100 = 22%.
Scenario 3:
Ka = 0.11
Initial concentration of acid (HA) = 0.150 M
By solving the equations as mentioned earlier, you would find that x = 0.0017 M.
Thus, the concentration of H+ = 0.0017 M.
Using the equation pH = -log[H+], we find that pH ≈ 2.77.
To find the percent dissociation, percent dissociation = (0.0017 M/0.150 M) * 100 = 1.13%.
I hope this explanation helps you understand how to solve these problems! Let me know if you have any more questions.