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Solubility product constant
Posted: Wed May 25, 2022 9:43 pm
by Dhamnekar Winod
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?
Re: Solubility product constant
Posted: Wed May 25, 2022 11:40 pm
by ChenBeier
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.
Re: Solubility product constant
Posted: Wed May 25, 2022 11:49 pm
by Dhamnekar Winod
I am working on a) and b).
(Deviation from the subject)What is PN? Would you explain?
Re: Solubiity product constant
Posted: Thu May 26, 2022 12:39 am
by ChenBeier
PN = Personal notification also PM Personal message or easier called email.
Re: Solubility product constant
Posted: Fri May 27, 2022 4:21 am
by alexshokz
Nice post
Re: Solubility product constant
Posted: Thu Aug 18, 2022 9:06 am
by Dhamnekar Winod
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)?
Re: Solubility product constant
Posted: Thu Aug 18, 2022 11:51 am
by ChenBeier
For Ag + 2 NH
3 => [Ag (NH
3)
2]+ + e- is E°= 0,413 V
Calculation here but it is in german
https://www.chemieunterricht.de/dc2/kom ... -best.html
Re: Solubility product constant
Posted: Thu Aug 18, 2022 10:58 pm
by Dhamnekar Winod
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?
Re: Solubility product constant
Posted: Fri Aug 19, 2022 6:30 am
by ChenBeier
In the link they also mentioned 1.2 x 10^7 . What is more close to your result with 1.9 x 10 ^7.