Chemistry Forum

a)Calculate the solubility product constant for Mg(OH)₂
Mg(OH)₂ + 2 e⁻ → Mg + 2OH⁻ E° = -2.67

b) Calculate the formation constant for Ag(NH₃)₂⁺

How to answer both these questions?

(Deviation from the subject) Why don't other moderators answer the questions asked by chemistry students, chemistry professionals, chemistry hobbyst in this forum?

1. Do you know the law of mass Action?
2. What is your own atempt to this.
3. Your deviation question, maybe send them a PN, here is not much traffic.

I am working on a) and b).

(Deviation from the subject)What is PN? Would you explain?

PN = Personal notification also PM Personal message or easier called email.

Nice post

My answer to (a):
\(Mg(OH)_2 + 2 e^- \leftrightarrow Mg + 2 OH^- E^{\circ} = -2.67 V \) oxidation

\( Mg + 2 e^- \rightarrow Mg E^{\circ}= -2.36 V K = \frac{1}{K_{sp}}\) reduction

\( log K = \frac{2 [-2.36 -(-2.67)]}{0.059} =10.51 \)

\(K=\frac{1}{K_{sp}}= 3.2 \times 10^{10} \)

\(K_{sp}\) = 3.1e-11

My answer to (b):

\(Ag + 2 NH_3 \rightarrow Ag(NH_3)_2^+ E^{\circ} =?\)

\(Ag^+ + e^- \rightarrow Ag E^{\circ} = 0.80 V \)

Now, how to answer (b)?

For Ag + 2 NH₃ => [Ag (NH₃)₂]+ + e- is E°= 0,413 V

Calculation here but it is in german

https://www.chemieunterricht.de/dc2/kom ... -best.html

I found on Internet, E°= 0.37 Volts for the reaction [Ag(NH₃)₂]⁺ + e⁻ ⇌ Ag + 2 NH₃.

So, K = Kf, Log Kf= \(\frac{0.80 Volts-0.37 Volts}{0.059} = 7.29 K_f= 1.9 \times 10^7 \)

There is a discrepancy between your answer of 2.1e6 and my answer 1.9e7. Is this discrepancy significant?

In the link they also mentioned 1.2 x 10^7 . What is more close to your result with 1.9 x 10 ^7.