0
$\begingroup$

I have a homework problem as follows:

A solution was prepared by mixing $25.0\ \mathrm{mL}$ of $0.208\ \mathrm{M}$ $\ce{Fe(NO3)3}$ and $4.50\ \mathrm{mL}$ of $0.00190\ \mathrm{M}$ $\ce{KSCN}$ and diluting to $50.00\ \mathrm{mL}$ with solvent, $0.10\ \mathrm{M}$ $\ce{HNO3}$. Calculate the initial concentrations of the thiocyanate ion and the iron(III) ion before any reaction takes place.

Now, how do I proceed with the question? At first I thought that it should just be the same concentration as the solutions, $0.208\ \mathrm{M}$ and $0.00190\ \mathrm{M}$, because there is only one thiocyanate and one iron(III) ion in the solutions. But the fact that we have an $\ce{HNO3}$ solution confuses me. Am I somehow supposed to use $M_1 V_1 = M_2 V_2$?

$\endgroup$
  • $\begingroup$ Welcome to Chemistry.SE. Take the tour to get familiar with this site. This appears to be a homework question, please share your thoughts and attempts towards the solution. It'll make us certain that ‎we aren't doing your homework for you. $\endgroup$ – user15489 Aug 29 '15 at 23:38
  • $\begingroup$ Nitric acid is like the question said just the solvent, so you are on the right track with your formula, as it states before any reaction. $\endgroup$ – Martin - マーチン Aug 30 '15 at 10:50
2
$\begingroup$

Am I somehow supposed to use $M_1V_1=M_2V_2$?

Yep!

The problem has given you three different solutions:

  1. Iron(III) nitrate
  2. Potassium thiocyanate
  3. Nitric acid

You know the concentrations of each, and you know (or can calculate) the volumes of each that are mixed to yield the final solution. What you'll actually be using in the calculation are three different equations:

  1. $M_{1i}V_1 = M_{1f}V_f$
  2. $M_{2i}V_2 = M_{2f}V_f$
  3. $M_{3i}V_3 = M_{3f}V_f$

$M_{\#i}$ is the initial concentration of each of the three chemicals in their separate starting solutions, and $M_{\#f}$ is the final concentration of each after mixing, that the problem is asking for (except for the $\ce{HNO3}$).

|improve this answer|||||
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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