Hello! It's me again.
How do I calculate % ionization for
a.) .50M HONH2 Kb = 1.1e-8
b.) .10M CH3NH2 kB = 4.38e-4
What is [OH-] and pH of these solutions
a.) 0.090M NaOH and 0.090M LiOH
b.) 0.050M Ba(OH)2 and 0.050M RbOH
a)
NH2OH ==> NH2^+ + OH^-
Kb = (NH2^+)(OH^-)/(NH2OH)
Start with 0.50 M NH2OH.
x M ionizes to give x M NH2^+ and x M OH^-. That leaves (NH2OH) = 0.50 M - x.
Plug in the numbers and calculate OH^-
You can solve the quadratic exactly BUT you can assume 0.50 - x = 0.50 (because x is small) and avoid the quadratic equation which will save some grinding of teeth.
Then percent ion = [(OH^-)/0.50]*100
b) is done the same way.
For the second set, both NaOH and LiOH are strong bases. Calculate OH^- from each, add them together, and determine pH and pOH
Same for Ba(OH)2 and RbOH; however, remember that the (OH^-) for Ba(OH)2 is twice the [Ba(OH)2].
Post your work if you want me to check it.
For a.) NH2OH
x^2/.50 = 1.1 x 10^-8 =>
.50(1.1 x 10^-8) = x^2
5.5 x 10^-9 = x^2
5.5 x 10^-9 square root = 7.416 x 10^-5
= x
these are all the steps I have
You are ok as far as you went except I would round the answer off to 7.4 x 10^-5 or carry one more through until the end, then round to 2 figures.
percent ionization = [(OH^-)/0.5]*100
Hello again! I'm glad you're back with more questions. Let's go through each question step-by-step.
a.) To calculate the percent ionization of a weak base, you need to know its concentration (M) and its base dissociation constant (Kb). The percent ionization can be calculated using the formula:
% ionization = [OH-] / initial concentration of weak base × 100%
In this case, you have a weak base HONH2 with a concentration of 0.50 M and a Kb value of 1.1 × 10^-8. To find the percent ionization, follow these steps:
1. Let x be the amount of HONH2 that ionizes. This can be assumed to be small compared to the initial concentration of the weak base.
2. The concentration of OH- formed is then x. Since HONH2 is a base, OH- is produced when it ionizes.
3. The reaction can be written as: HONH2 + H2O ⇌ ONH- + H3O+
4. From the equation, the concentration of HONH2 remaining will be (0.50 - x).
5. Using the Kb expression: Kb = [ONH-][H3O+] / [HONH2], you can write Kb = x^2 / (0.50 - x).
6. Solve the quadratic equation for x: Kb = x^2 / (0.50 - x), where x is the concentration of OH-. Solve for x and find its value.
7. Finally, calculate the percent ionization using the formula mentioned earlier.
b.) Similarly for the weak base CH3NH2 with a concentration of 0.10 M and a Kb value of 4.38 × 10^-4, follow the same steps to calculate percent ionization.
Now, let's move on to the second part of your question.
a.) To find the [OH-] and pH of a solution, you need to know its concentration. In this case, you have 0.090 M NaOH and 0.090 M LiOH.
1. Since NaOH and LiOH are strong bases, they completely dissociate in water to give OH- ions.
2. Therefore, the concentration of OH- in both solutions is equal to the concentration of the strong base.
3. So, [OH-] for both NaOH and LiOH is 0.090 M.
4. To find the pH, you can use the equation: pH = -log[H+]. Since the OH- concentration is known, you can use the relation [H+][OH-] = Kw. Kw is the ionization constant for water, which is 1.0 × 10^-14 at 25°C.
5. From [H+][OH-] = Kw, you can solve for [H+], and then find the pH using -log[H+].
b.) Similar to the previous case, for 0.050 M Ba(OH)2 and 0.050 M RbOH, the concentration of [OH-] is equal to their respective concentrations of the strong bases. To find the pH, apply the same steps as mentioned above.
Remember to use appropriate units in your calculations. I hope this explanation helps! Let me know if you have any further questions.