What concentration of Ag+ is in equilibrium with 3.0×10−6 M Co(CN)3−6 and Ag3Co(CN)6(s)?
Ag3Co(CN)6 --> 3Ag^+3 + Co(CN)6 ^-3
I worked the problem as such:
Ksp= 3.9*10^-26
3.9*10^-26= [3x]^3 * [x]
3.9*10^-26= [3x]^3 * [3.0*10^-6]
Divided both sides by 3.0*10^-6 to isolate the [3x]^3, which yields:
1.3*10^-20=[3x]^3
Divided both sides by 3 to isolate [x]^3, which yields:
4.33*10^-21=[x]^3
Took the cube root of both sides to isolate x, which yields:
1.63*10^-7=x
Yet this answer is apparently wrong. Can you help me?
1.3*10^-20=[3x]^3
Divided both sides by 3 to isolate [x]^3, which yields:
4.33*10^-21=[x]^3
See your work above. That term of (3X)^3 is 27X^3 so the next step is to divide by 27 and not 3. I didn't check anything after that.
Thank you. Unfortunately I still got an incorrect answer after doing:
1.3*10^-20=[3x]^3 divide both sides by 27, which yields:
4.81*10^-22=[x]^3 then I took the cubed root (3√(4.81*10^-22)
Which yields: 7.835*10^-8=x
Thank you very much for your time and help! I appreciate it so much.
It may be worthwhile to check your answer again with significant figures. The use of 3.9E-26 limits the number of s.f. to 2 so
the answer should be reported as 7.8E-8 M. Some online databases will count too many significant figures as an incorrect answer.
Your approach to the problem is correct, but there is a mistake in one of your calculations.
Let's go through it step by step:
1. Write the balanced equation for the dissolution of Ag3Co(CN)6:
Ag3Co(CN)6(s) ⇌ 3 Ag+(aq) + Co(CN)6^3-(aq)
2. Write the expression for the solubility product constant, Ksp:
Ksp = [Ag+]^3 [Co(CN)6^3-]
3. Plug in the given values:
Ksp = (3x)^3 (3.0 × 10^(-6))
4. Simplify:
Ksp = 27x^3 (3.0 × 10^(-6))
5. Divide both sides by 27 to isolate x^3:
Ksp/27 = x^3 (3.0 × 10^(-6))
6. Now, here's where the error occurred. You divided the entire equation by (3.0 × 10^(-6)), but forgot to divide Ksp/27 as well. So, make sure to divide Ksp/27 by (3.0 × 10^(-6)):
(Ksp/27) / (3.0 × 10^(-6)) = x^3
7. Simplify:
Ksp / (81 × 10^(-6)) = x^3
8. Calculate the numerical value of Ksp / (81 × 10^(-6)):
Ksp / (81 × 10^(-6)) = 4.81 × 10^(-21)
9. Now, take the cube root of both sides to solve for x:
x = (4.81 × 10^(-21))^(1/3) ≈ 7.09 × 10^(-7)
Therefore, the concentration of Ag+ in equilibrium with 3.0 × 10^(-6) M Co(CN)3^(-6) and Ag3Co(CN)6(s) is approximately 7.09 × 10^(-7) M.