Solubility product constant

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Dhamnekar Winod
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Solubility product constant

Post 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?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Solubility product constant

Post 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.
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Re: Solubility product constant

Post by Dhamnekar Winod »

I am working on a) and b).

(Deviation from the subject)What is PN? Would you explain?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Solubiity product constant

Post by ChenBeier »

PN = Personal notification also PM Personal message or easier called email.
alexshokz
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Re: Solubility product constant

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Nice post
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Dhamnekar Winod
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Re: Solubility product constant

Post 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)?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Solubility product constant

Post by ChenBeier »

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
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Re: Solubility product constant

Post 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?
Any science consists of the following process. 1) See 2)Hear 3)Smell,if needed 4)Taste, if needed 5)Think 6)Understand 7)Inference 8)Take decision [Believe or disbelieve, useful or useless, true or false, cause or effect, any other criteria]
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Re: Solubility product constant

Post 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.
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