3
$\begingroup$

Recently I've been looking at this website, which says:

The $K_{\mathrm{sp}}$ of $\ce{Cr(OH)3}$ is $6.70 \times 10^{-31}$

Problem 1: What is the minimum $\mathrm{pH}$ at which $\ce{Cr(OH)3}$ will precipitate?

$$\ce{Cr(OH)3\rightleftharpoons {Cr^{3+}}+3OH-}$$ $$K_{\mathrm{\mathrm{sp}}}=[\ce{Cr^{3+}}][\ce{OH-}]^{3}$$ $$[\ce{Cr^{3+}}]=s$$ $$[\ce{OH^{-}}]=3s$$ Problem 2: What is the minimum $\mathrm{pH}$ at which $\ce{Cr(OH)3}$ will precipitate if the solution has $[\ce{Cr^{3+}}] = 0.0670~\mathrm{M}$? $$\ce{Cr(OH)3\rightleftharpoons {Cr^{3+}}+3OH-}$$ $$K_{\mathrm{sp}}=[\ce{Cr^{3+}}][\ce{OH-}]^{3}$$ $$6.70 \times 10^{-31}=[0.0670~\mathrm{M}][s^{3}]$$

I don't understand why the site wrote this:

Note that $3s$ is not necessary. In example one, $s$ was assigned to the chromium ion and we knew the hydroxide to be three times greater than that value. Here, the hydroxide is simply an unknown value and it is not expressed in terms of some other unknown value (as it was in Problem 1).

Can anyone tell me why?

Also is also used to other equilibrium constant equation? What did I miss?

Improve this question
$\endgroup$
0

1 Answer 1

Reset to default
0
$\begingroup$

Let the concentration of $\ce{Cr^{3+}}$ be $t$ then the concentration of $\ce{OH-}$ will be $3t$.

So, the site says that writing the concentration as $3t$ is not necessary.

What it is basically doing is considering $3t$ equal to some variable $s$, and solving the question. There is nothing wrong in this, is it?

Now, for example we had not done that and calculated the value of $t$ by using the equation

$K_{\mathrm{sp}}=[\ce{Cr^{3+}}] \cdot (3t)^3$

You will then multiply the derived value of $t$ with 3 to get $\ce{[OH-]}$.

From which you will find the $\mathrm{pH}$. So the site says:

"Why not calculate the value of $3t$ directly by using a variable $s$?"

Therefore: $s=3t$

Therefore: $\mathrm{pOH}=-\log{s}$

What you are thinking is also correct, and the answer with that will be: $\mathrm{pOH}=-\log{3t}$

So you both are doing the exact same thing here.

Improve this answer
$\endgroup$
8
  • $\begingroup$ I'm sorry, but I can't get it. Is using 3s just more good than making new variable (3t)? Could you provide full answer? $\endgroup$
    – làntèrn
    May 10, 2015 at 3:43
  • $\begingroup$ @Ridho wherever you used s in the above question, I have used t. Just to make you clear on the concept. Here, 3t=s in the question. Here I am using concentration as t and not s, which we usually do. What the site means to say is that you need the vale of 3t, so why do we need to first find the value of t and multiply it be 3. In short it did that directly $\endgroup$
    – Rajat Jain
    May 10, 2015 at 3:45
  • $\begingroup$ @Ridho I have appended the answer now. The site has just replaced your $3t$ to an independent variable $s$. $\endgroup$
    – Rajat Jain
    May 10, 2015 at 3:50
  • $\begingroup$ Just forget all about 3s, All the s variables above have been replaced by t variables by me. $\endgroup$
    – Rajat Jain
    May 10, 2015 at 3:58
  • $\begingroup$ So I plug the t like this: $6.70\cdot10^{-31}=(0.0670\:M)\cdot(3\cdot0.0670\:M)^{3}$? $\endgroup$
    – làntèrn
    May 10, 2015 at 4:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.