i need help.
Calculate the pH of a 0.24 M CoCl3 solution. The Ka value for Co(H2O)63+ is 1.0 10-5.
how do u do these with like the Cl there?
Responses
Chemistry - DrBob222, Monday, February 16, 2009 at 12:35am
Just ignore the Cl^-.
Co(H2O)6Cl3 ==> Co(H2O)6^+3 + 3Cl^-
Then the hexaaquocobalt(III) ion ionizes as (and I will write this a funny way to help you see it).
Co(H2O)6^+3 ==>Co(H2O)5(H2O)^+3==> Co(H2O)5(OH)^+2 + H^+.
Then you write the Ka expression for that last part and solve for H^+ after doing an ICE chart.
i did wat dr. bob said to do and i got a pH of 1.31
what did i do wrong?
Well, it seems like your calculations went a bit sour! Don't worry, I'm here to help you out. Let's see what might have gone wrong.
First, let's take a look at Dr. Bob's explanation. He suggested that you ignore the chloride ion (Cl^-) and focus on the equilibria involving the hexaaquocobalt(III) ion (Co(H2O)6^+3).
Based on Dr. Bob's response, the hexaaquocobalt(III) ion can further ionize to form:
Co(H2O)6^+3 => Co(H2O)5(H2O)^+3 => Co(H2O)5(OH)^+2 + H^+
Now, you need to write the Ka expression for this last part and solve for the concentration of H^+(hydrogen ions) after making an ICE chart.
That being said, I can't really tell where the mistake happened without seeing your working. My suggestion would be to double-check your calculations and make sure that you're plugging in the correct values into your equations. One small slip-up can really throw off the pH calculation.
If you're still having trouble, you can share your calculations with me, and I'll be more than happy to take a look. Just remember, even in chemistry, sometimes mistakes happen. But hey, at least with humor, you can shake off those errors and get back on track!
Based on the information provided, it seems like you followed the correct steps, but there may have been a mistake in your calculation. To verify your pH value, let's go through the steps again step-by-step.
1. Write the dissociation reaction for Co(H2O)6^+3:
Co(H2O)6^+3 ⇌ Co(H2O)5(OH)^+2 + H^+
2. Write the expression for the equilibrium constant (Ka) for this reaction:
Ka = [Co(H2O)5(OH)^+2] [H^+] / [Co(H2O)6^+3]
3. Set up an ICE (initial, change, equilibrium) table:
Initial: [Co(H2O)6^+3] = 0.24 M, [Co(H2O)5(OH)^+2] = 0 M, [H^+] = 0 M
Change: -x, +x, +x
Equilibrium: 0.24 - x, x, x
4. Substitute the equilibrium concentrations into the Ka expression:
Ka = x^2 / (0.24 - x)
5. Assume x is small compared to 0.24, which is reasonable since x represents the concentration of H^+ ions generated from the reaction.
√(Ka * (0.24 - x) ≈ x
This simplifies the equation to:
x = √(Ka * (0.24))
6. Calculate x using the given Ka value (Ka = 1.0 * 10^-5) and initial concentration (0.24 M):
x = √(1.0 * 10^-5 * 0.24)
x ≈ 4.9 * 10^-4 M
7. Calculate the pH using the concentration of H^+:
pH = -log[H^+]
pH = -log(4.9 * 10^-4)
pH ≈ 3.31
So, the correct pH of the 0.24 M CoCl3 solution should be approximately 3.31, not 1.31.
Based on the information provided by Dr. Bob, it seems like you followed the correct steps. However, calculating the pH of a solution involves a few additional calculations. Let me explain the process in more detail:
1. Start with the equation: Co(H2O)6Cl3 → Co(H2O)6^3+ + 3Cl^-
In this equation, the hexaaquocobalt(III) chloride dissociates to form the hexaaquocobalt(III) ion and 3 chloride ions.
2. Next, the hexaaquocobalt(III) ion can further dissociate:
Co(H2O)6^3+ → Co(H2O)5(H2O)^2+ + H+
This means that one water molecule is replaced by a hydroxide ion (OH-) and a hydrogen ion (H+).
3. Write the Ka expression for this dissociation:
Ka = [Co(H2O)5(OH)]^2+[H+] / [Co(H2O)6^3+]
The brackets [ ] indicate the concentration of each species.
4. Set up an ICE (Initial, Change, Equilibrium) chart to solve for the concentration of H+:
I: [Co(H2O)6^3+] [Co(H2O)5(OH)]^- [H+]
C: -x +x +x
E: [Co(H2O)6^3+ - x] [Co(H2O)5(OH)]^- + x [H+]
5. Substitute the equilibrium concentrations into the Ka expression:
Ka = (x)(x) / (0.24 - x)
6. Since the concentration of H+ is very small compared to 0.24, we can simplify the expression to:
Ka ≈ x^2 / 0.24
7. Use the given Ka value (1.0 x 10^-5) to solve for x:
Ka = x^2 / 0.24
1.0 x 10^-5 = x^2 / 0.24
x^2 = 1.0 x 10^-5 * 0.24
x = √(1.0 x 10^-5 * 0.24)
8. Calculate the pH using the concentration of H+:
pH = -log[H+]
Make sure to plug in the correct values and check your calculations. If you followed these steps and still obtained a different result, please double-check your math or provide more details on your calculations for further assistance.