I have a hard time understanding equilibrium problems...

[Temperature101]

The K

sp of Mg(OH)2 in water is 1.2 × 10^-11

​

​

mol​

​

3/L3. If the Mg2+ concentration in an acid

solution is 1.2 ​

​

× 10^–5, what is the pH at which

Mg(OH)​

​

2 just begins to precipitate?

​

​

A . ​

​

3

​

​

B . ​

​

4

​

​

C . ​

​

5
D . 11

The answer is D...Can someone explain?

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[indianjatt]

First of all, you want to determine at concentration of OH- will Mg(OH)2 precipitate if the concentration of Mg is 1.2 x 10^-5 and Ksp = 1.2 x 10^-11

So:
Ksp= [Mg2+][OH-]^2
1.2x10^-11/1.2x10^-5=[OH-]^2
[OH-]=1x 10^-3
POH = 3
PH + POH = 14
PH = 11.

Answer is D.

 

[Temperature101]

First of all, you want to determine at concentration of OH- will Mg(OH)2 precipitate if the concentration of Mg is 1.2 x 10^-5 and Ksp = 1.2 x 10^-11

So:
Ksp= [Mg2+][OH-]^2
1.2x10^-11/1.2x10^-5=[OH-]^2
[OH-]=1x 10^-3
POH = 3
PH + POH = 14
PH = 11.

Answer is D.

Thanks...Equilibrium is so complex; therefore, simple problem can sometimes appear complicated.

 

[salim271]

Thanks...Equilibrium is so complex; therefore, simple problem can sometimes appear complicated.

Best to keep in mind that although I am a firm believer than MCAT test writers are evil to the core, that they have to stick to the rules and write tests where questions can be answered in at least 3 minutes or less without a calculator. There usually is a simple-ish solution to every MCAT problem, the biggest trick is knowing the process, or figuring out the process quickly.

 

[Temperature101]

Best to keep in mind that although I am a firm believer than MCAT test writers are evil to the core, that they have to stick to the rules and write tests where questions can be answered in at least 3 minutes or less without a calculator. There usually is a simple-ish solution to every MCAT problem, the biggest trick is knowing the process, or figuring out the process quickly.

Yeah... I wish they would ask these questions in a simple way. For the problem I posted, if they had asked: What is the PH of the solution? I would have answered it. However, they ask for the PH in a convoluted way...I wish they stop doing that...