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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: nacon2012 on April 25, 2013, 02:56:54 PM

Title: Modified Solubility Product Constant, Ksp for Mg(OH)2 . . .
Post by: nacon2012 on April 25, 2013, 02:56:54 PM

When excess solid Mg(OH)₂ is shaken with 1.00 L of 1.0 M NH₄Cl solution, the resulting saturated solution has pH = 9.0. Calculate K_sp of Mg(OH)₂.

The book gives a value of 1.8*10^-11 = K_sp

No other information is provided; However, Usually, for Mg(OH)₂, K_sp = 4*S³. So, that would mean, that S=.000165096 per book answer.

The book also gives a value at standard conditions for K_sp of Mg(OH)₂ = 2.06E-13. This would give, S=0.0000372051. Which in turn would be pH=9.87, I believe, please check all of these values as you perform your procedure. . . The was a HW problem which is already late, I am very interested in the procedure to solve this problem. Thank You very much; please do not send any information unless you know your procedure leads to the correct answer, I have tried numerous systematic procedures to get a value near the book value, and to no avail.

Title: Re: Modified Solubility Product Constant, Ksp for Mg(OH)2 . . .
Post by: Borek on April 25, 2013, 03:57:42 PM

Hint: how does ratio [NH₃]/NH₄⁺] depend on pH?

Title: Re: Modified Solubility Product Constant, Ksp for Mg(OH)2 . . .
Post by: nacon2012 on April 26, 2013, 06:41:38 PM

Quote from: Borek on April 25, 2013, 03:57:42 PM

Hint: how does ratio [NH₃]/NH₄⁺] depend on pH?

pH=pK_a+Log[A^-/HA] as pH increases so too does the log[NH3]/NH4+] increases . . .

I am unable to continue this.

If OH is used to react with ammonium, then, The original OH can not contribute to pH directly, unless the Ammonium + OH reaction is 1 to 1, instead of 2 to 2. This was a challenge problem in our homework assignment . . .

All of this to no avail. The Produced Ammonia may contribute to pH, but if only 1 mol of OH (of the 2 mol produced from our original Metal) reacts with Ammonium, then that other mol of OH (that didn't react with ammonium) will suppress the production of OH from NH3. So, likely it is a 1 to 1 mol ratio, and therefore only 1 mol of OH (from the original 2 mols) can contribute to the pH. I have tried this . . . I have reviewed my algebra on several occasions. Hints are helpful . . .

I do appreciate the hint, but keep them coming if you will. Thank You.

Title: Re: Modified Solubility Product Constant, Ksp for Mg(OH)2 . . .
Post by: Borek on April 27, 2013, 02:39:42 AM

I am not sure I follow what you wrote, as I have no idea if/when you are talking about ammonia (NH₃) and ammonium (NH₄⁺).

You are right about the ratio. This is half of the solution.

The other half: where does ammonia (NH₃) in the solution comes from?