What is the molarity of a solution composed of 5.00*g sodium sulfate dissolved in 125*mL of solution?

Oct 7, 2017

We use the quotient....$\text{Concentration"="Moles of solute"/"Volume of solution}$

Explanation:

And such a quotient gives us an answer with units of $m o l \cdot {L}^{-} 1$...and thus...

$\left[N {a}_{2} S {O}_{4}\right] = \frac{\frac{5.00 \cdot g}{142.04 \cdot g \cdot m o {l}^{-} 1}}{125 \cdot m L \times {10}^{-} 3 \cdot L \cdot m {L}^{-} 1}$

$= 0.282 \cdot m o l \cdot {L}^{-} 1$ with respect to $\text{sodium sulfate}$....

And we know that sodium sulfate speciates in aqueous solution according to the reaction....

$N {a}_{2} S {O}_{4} \left(s\right) \stackrel{{H}_{2} O}{\rightarrow} 2 N {a}^{+} + S {O}_{4}^{2 -}$

And so $\left[N {a}^{+}\right] = 0.563 \cdot m o l \cdot {L}^{-} 1$, and $\left[S {O}_{4}^{2 -}\right] = 0.282 \cdot m o l \cdot {L}^{-} 1$.

Do you agree?

Oct 7, 2017

Molarity of the solution: 0.12

Molarity of sulfate ion: 0.360

Molarity of aluminum ion: 0.240

Explanation:

So first, it would be helpful to write the molecular formula of aluminum sulfate. Since you're working with an ionic compound, the goal is to make the separate charges of aluminum and sulfate equal zero.

Aluminum's charge is $A {l}^{3 +}$. Aluminum sulfate's charge is $S {O}_{4}^{2 -}$. To balance the two, let's multiply the cation by $2$ - that will give us a $+ 6$ charge. And let's multiply the anion by $3$ - that will give us a $- 6$ charge. So the molecular formula for aluminum sulfate is: $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$. Notice that the charges cancel, which is why they're not in the molecular formula anymore!

Now let's actually answer the first part of the question. Molarity is equal to the number of moles of a substance over the amount of liters. We are given grams of aluminum sulfate, which we can convert to moles. Let's do that first:

$5.00$ grams $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$ = $\left(1 \text{ mole ")/(342.15 " g } A {l}_{2} {\left(S {O}_{4}\right)}_{3}\right)$ $= 0.015$ moles $A {l}_{2} {\left(S {O}_{4}\right)}_{3}$

342.15 grams is the molar mass of aluminum sulfate. You can use this value to convert from moles to grams and from grams to moles, depending on what's given in the problem.

Now we have everything we need to calculate the molarity, but don't forget to convert mL into L:

Molarity = $\left(0.015 \text{ moles " Al_2(SO_4)_3) / (0.125 " L}\right) = 0.12$

That's the answer to your first problem; now let's look at the second part. Aluminum and sulfate, according to the molecular formula we found, is in a ratio of 2 to 3. This means that we can use stoichiometry to convert from moles of aluminum sulfate to aluminum and sulfate ions. That's all the information needed to calculate the molarity of each ion:

$0.015 \text{ moles " Al_2(SO_4)_3 * (3 " moles " SO_4)/(1 " mole " Al_2(SO_4)_3) = 0.045 " moles } S {O}_{4}$

$0.015 \text{ moles " Al_2(SO_4)_3 * (2 " moles " AL_2)/(1 " mole " Al_2(SO_4)_3) = 0.030 " moles } A {l}_{2}$

Now we can calculate the molarity:

Molarity of $S {O}_{4}$ = $\frac{0.045 \text{ moles } S {O}_{4}}{0.125 L} = 0.360$

Molarity of $A {l}_{2}$ = $\frac{0.030 \text{ moles } S {O}_{4}}{0.125 L} = 0.240$