Here is my response to you earlier question. If you still have questions you may want to follow up here.
You didn't substitute correctly for Keq. You should calculate
log Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+)
You made log Keq = 0.27. You should follow the instructions I gave.
(Pb) = 1
(Ni^2+) = 0.27
(Ni) = 1
(Pb^2+) = x so the equation is
0.020 = 0.12 - (0.0296)*log[(1)(0.27)/(1)(x)] and solve for x in moles/L.
I don't want to leave without acknowledging an error in what I wrote. It make no difference in the final answer becasue I corrected it later. Here is what I wrote for ONE of the steps.
Ecell = Eo cell - (2.303*RT/n) log Q and Q = Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+) but I omitted Faraday's constant. It should be
Ecell = Eo cell - (2.303*RT/nF) log Q and Q = Keq=(Pb)(Ni^2+)/(Ni)(Pb^2+)
Then if you substitute for R of 8.314, T = 298, F = 96,485 and 2 for n, then
(2.303*8.314*298/2*96,485) = 0.0296. Your post is now down below the line. You may want to repost at the top if you have additional quesitons.
Thank you for your response and clarification. Let's address your calculations and understand how to correctly substitute for Keq and solve for x.
In your initial equation, you have:
log Keq = (Pb)(Ni^2+)/(Ni)(Pb^2+)
To solve for x in moles/L, you need to substitute the appropriate values into the equation:
0.020 = 0.12 - (0.0296)*log[(1)(0.27)/(1)(x)]
Now, let's break down the steps on how to solve this equation:
1. Start with the given equation:
0.020 = 0.12 - (0.0296)*log[(1)(0.27)/(1)(x)]
2. Simplify the expression within the log:
0.020 = 0.12 - (0.0296)*log[0.27/x]
3. Move the constant term (0.12) to the other side of the equation:
(0.12 - 0.020) = (0.0296)*log[0.27/x]
0.1 = (0.0296)*log[0.27/x]
4. Divide both sides by 0.0296 to isolate the log term:
0.1 / 0.0296 = log[0.27/x]
3.378 = log[0.27/x]
5. Rewrite the equation using exponentiation:
10^3.378 = 0.27/x
2174.43 = 0.27/x
6. Cross-multiply to solve for x:
x = 0.27/2174.43
x = 0.000124 moles/L (rounded to the appropriate number of significant figures)
So, the value of x is approximately 0.000124 moles/L.
I hope this clarifies the process of substituting for Keq and finding the value of x in moles/L. Let me know if you have any further questions or if there's anything else I can assist you with!