## Solubility product constant

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Dhamnekar Winod
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### Solubility product constant

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]
ChenBeier
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### Re: Solubility product constant

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.
Dhamnekar Winod
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Posts: 243
Joined: Sat Nov 21, 2020 10:14 am
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### Re: Solubility product constant

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]
ChenBeier
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### Re: Solubiity product constant

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

Nice post
Dhamnekar Winod
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### Re: Solubility product constant

$$Mg(OH)_2 + 2 e^- \leftrightarrow Mg + 2 OH^- E^{\circ} = -2.67 V$$ 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

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

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

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]
ChenBeier
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### Re: Solubility product constant

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
Dhamnekar Winod
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### Re: Solubility product constant

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]
ChenBeier
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### Re: Solubility product constant

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