# Given solubility of sodium carbonate, find mass of hydrate sodium carbonate dissolved?

I am working on a problem from my scholarship exam practice (MEXT). I am not quite sure if I got it right. The question was provided with multiple choices (a) to (f) below:

The solubility of sodium carbonate in $$\pu{100 g}$$ water is $$\pu{25.0 g}$$ at $$\pu{22 ^{\circ}C}$$. How many grams of the hydrate $$\ce{Na2CO3∙10H2O}$$ can be dissolved in $$\pu{100 g}$$ of water at $$\pu{22 ^{\circ}C}$$? (a) $$\pu{0.556g}$$; (b) $$\pu{0.762g}$$; (c) $$\pu{9.27g}$$; (d) $$\pu{67.5g}$$; (e) $$\pu{81.7g}$$; (f) $$\pu{117g}$$

I honestly do not know how to solve this problem. My best guess is to find the moles of sodium carbonate that can be dissolved in water first (from $$\pu{25.0 g}$$). And use that moles equate to the moles of hydrate form (1:1). Then find the mass of hydrate form by multiplying its molecular weight and my answer would be $$\pu{67.5g}$$. There is no answer key provided so I would like to hear other opinions and get some advice.

To answer this kind of problem, it'd be better starting with elimination approach. It is given the solubility of sodium carbonate in $$\pu{100 g}$$ of water as $$\pu{25.0 g}$$ at $$\pu{22 ^{\circ}C}$$. By that, you know, solubility of its hydrate should be more than $$\pu{25.0 g}$$ at $$\pu{22 ^{\circ}C}$$ because the mass of hydrate is including additional mass of water as well (as your calculation shows). Therefore, you can eliminate answers (a), (b), and (c) directly, because they are less than 25 and are not correct. Your calculations shows it must be greater than $$\pu{67.5 g}$$ (therefore, answer (d) is incorrect as well). It should be greater than $$\pu{67.5 g}$$, because you forgot to include additional amount of salt, which would dissolve in 10 equivalents of water in that added salt: Molar mass of $$\ce{Na2CO3}$$ is $$\pu{106 gmol^{-1}}$$ while Molar mass of $$\ce{Na2CO3.10H2O}$$ is $$\pu{286 gmol^{-1}}$$. This means, each $$\pu{mol}$$ of hydrate has $$\pu{186 g}$$ or $$\pu{186 mL}$$ of water. That is a lot and we need to consider that as well in this question. Now we have only two answers to be considered: (e) and (f).

To do the calculations, suppose $$x\; \pu{g}$$ of $$\ce{Na2CO3.10H2O}$$ would dissolve in $$\pu{100 g}$$ of water at $$\pu{22 ^{\circ}C}$$. Thus, $$\text{amount of }\ce{Na2CO3}\text{ in the solution} = \left(\frac{\pu{106 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$ $$\text{and, amount of }\ce{H2O}\text{ in the solution} = \pu{100 g} + \left(\frac{\pu{10 \times 18 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$ From given data for anhydrous salt, we can conclude that upon saturation: $$\text{The amount of }\ce{H2O}\text{ in the solution} = 4\times \text{amount of }\ce{Na2CO3}\text{ in the solution}$$

$$\therefore \pu{100 g} + \left(\frac{\pu{10 \times 18 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g} = 4 \times \left(\frac{\pu{106 gmol^{-1}}}{\pu{286 gmol^{-1}}}\right) x \;\pu{g}$$ Solve for $$x$$: $$\left(4 \times 106 - 180\right)x = 28600 \; \text{and therfore, } x=\frac{28600}{244}=117.2$$ Therefore your answer is (f) ($$\pu{117 g}$$).

Your calculation would be good, if you did not forget that hydrates contain water.

The solubility of the anhydrous carbonate is 25 g in 100 g of water.

Both carbonate forms, once diluted, are the same compound, forming the same ions plus eventually releasing water.

So in case of the hydrate, the ratio carbonate:water must be 1:4 as well.

$$M_{\ce{Na2CO3}}=106\mathrm{~g/mol}$$.
$$M_{\ce{Na2CO3. 10 H2O}}=286\mathrm{~g/mol}$$.

Using $$x$$ grams of the hydrate + $$100$$ grams of water
produces solution of $$x \cdot 106/286 \mathrm{g}~\ce{Na2CO3}$$
in $$100 + x \cdot 180/286~ \mathrm{g}$$ of water.

I will leave the rest on you to give you the honour of some effort, as this question is rather the homework class of questions.

See Homework