# How do I calculate the molarity of the Na2CO3 solution?

My teacher isn't very good at helping me solve this problem. I keep getting the answer, but I don't know why I am getting it. Could you please help me understand this problem and why you should do this specific step?

A $$\pu{1.00 g}$$ sample of $$\ce{Na2CO3.10H2O}$$ was dissolved in $$\pu{20.0 mL}$$ of distilled water. Additional water was added so as to give $$\pu{250 mL}$$ of solution. What is the molar concentration of $$\ce{Na2CO3}$$?

Here is what I did and got the answer and don't understand why I need to do it:

\begin{align} \frac{\pu{1 g}~\ce{Na2CO3.10H2O}}{\pu{286 g}~\ce{Na2CO3.10H2O}} &= \pu{0.003447 mol}\\ \frac{\pu{0.003447 mol}}{\pu{0.020 L}} &= \pu{0.174825 M}\\ (\pu{0.174825 M})(\pu{0.020 L}) &= (\pu{0.250 L})(x)\\ x &= \pu{0.013986 M} \end{align}

$$\pu{0.013986 M}$$ is the correct answer, but I don't know why and don't understand the question. If you were to do the problem, could you explain why you did what you did to get to the answer.

• @AwesomeFlame123 You clearly figured out the math aspect of the problem. Where is your issue, specifically? Which step is it that you are doing that you don't understand? You also don't need the dilution step at all. Once you've got moles, that can go directly into the molarity equation with 0.250L as your volume. None of the solute ever leaves the solution, after all. As written, you're dividing by 0.020 then immediately multiplying by 0.020. – Jason Patterson Oct 6 '14 at 3:18
• @JasonPatterson I was trying to just use the formula M1V1 = M2V2 – Asker123 Oct 6 '14 at 10:10
• Technically you don't know the volume of the solution of 20ml of water and sodium carbonate. You don't know a "true" volume until you dilute to 250 ml. – MaxW Oct 21 '15 at 16:33

What the question is asking is the molar concentration of the given substance after dilution with the help of values given. The first step of the answer is converting the given weight of $$\ce{Na2CO3.10H2O}$$ into amount of substance. For that we have a formula as \begin{align} \text{amount of substance} &= \frac{\text{mass of substance}}{\text{molecular mass of the substance}}\\ \text{amount of substance} &= \frac{ \pu{1 g}~\ce{Na2CO3.10H2O}}{\pu{286 g}~\ce{Na2CO3.10H2O}} = \pu{0.003447 mol}.\\ \end{align} We used this first step to determine the amount of substance of $$\ce{Na2CO3.10H2O}$$ present in the solution. It is necessary because even after dilution of the solution the amount of substance will remain constant.

The second step is converting the amount of substance into molarity. As, \begin{align} \text{molarity} &= \frac{\text{amount of substance}}{\text{volume in L}}\\ \text{molarity} &= \frac{\pu{0.003447 mol}}{\pu{0.020 L}} = \pu{0.174825 M}\\ \end{align}

The second step is done to find out the concentration of the substance in terms of molarity since the question has asked us to give the answer in molarity after dilution.

After that the given solution is diluted up to $$\pu{250 mL}$$, so we have to use dilution formula $$\begin{multline} \text{initial concentration}\times \text{initial volume of solution} =\\ \text{final concentration}\times \text{final volume of solution}. \end{multline}$$ So we suppose that the final concentration of the solution is $$x$$ and we have values for others entities. \begin{align} (\pu{0.174825 M})(\pu{0.020 L}) &= (\pu{0.250 L})(x)\\ x &= \pu{0.013986 M}\\ \end{align} This last step is carried out because after dilution the molarity of the solution changes but not the amount of substance as I stated earlier. So we equate the equation as $$\text{initial amount of substance} = \text{final amount of substance}.$$ And the final answer is $$\pu{0.013986 M}$$.

• Thank You! I didn't understand that the Molarity of the final Na2CO3 is the same as the molarity of Na2CO3⋅10H2O. – Asker123 Oct 6 '14 at 10:08
• Welcome. Well molarity is always calculated for the solution .So when crystals of $\ce{Na2CO3 * 10H2O}$ is dissolved in water then water of crystallization plays no role in determination of molarity since only no of moles of$\ce{Na2CO3}$ is taken in account. – CCR Oct 6 '14 at 11:21
• This is needlessly complicated. – Jan Feb 21 '20 at 8:27
• Please note that descriptive terms or names of quantities shall not be arranged in the form of an equation and that the quantity “amount of substance” shall not be called “number of moles”, just as the quantity “mass” shall not be called “number of kilograms”. – user7951 Feb 22 '20 at 12:40

The first part makes more sense writing moles explicitly. The first step is determining how many moles of sodium carbonate are in the original $$1.00\ \mathrm g$$ sample:

$$\frac{1.00\ \mathrm{g}}{286\ \mathrm{g/mol}} = 0.003447\ \mathrm{mol}$$

The second step is determining the concentration of the first $$20.0\ \mathrm{mL}$$ solution, i.e. what concentration results from dissolving the $$0.00345\ \mathrm{mol}$$ of sodium carbonate calculated above. $$\frac{0.003447\ \mathrm{mol}}{0.0200\ \mathrm{L}} = 0.174825\ \mathrm{M}$$

The third part is simply a dilution – what concentration results when we dilute the first $$20.0\ \mathrm{mL}$$ solution to $$250\ \mathrm{mL}$$: $$c_2 = \frac{c_1V_1}{V_2}$$ $$c_2 = \frac{(0.174825\ \mathrm{M})(0.020\ \mathrm{L})}{(0.250\ \mathrm{L})} = 0.0140\ \mathrm{M}$$

• No arguments at all about what you've written here, but to be clear finding the 20mL concentration and using the dilution equation is just extra work. Once we know the number of moles of sodium carbonate decahydrate, we can go straight to its molarity as the solute in 250mL of solution. – Jason Patterson Oct 6 '14 at 3:14
• You're quite right, but he asked for an explanation of the steps he took. – Michael DM Dryden Oct 6 '14 at 4:12
• Comments are not for extended discussion; this conversation has been moved to chat. – user7951 Feb 23 '20 at 13:53