Question 45

A packet of washing soda crystals has been left open to the atmosphere for some time. The crystals have formed a white powder which may be represented by the formula $$\mathrm{Na_2CO_3 \cdot xH_2O}$$. The value of $$x$$ may be found by dissolving a known quantity of the powder in water and titrating the solution with acid of known concentration.<br>In one experiment a student weighed out 1.59 g of the powder and dissolved it in water. The volume of the solution was made up to exactly $$250\ \text{cm}^3$$. $$25\ \text{cm}^3$$ of this solution was titrated with hydrochloric acid of concentration $$0.10\ \text{mol/dm}^3$$ (mol per litre). $$25.6\ \text{cm}^3$$ of the acid was required for complete reaction. The equation for this reaction is<br>$$\mathrm{Na_2CO_3 + 2HCl \rightarrow 2NaCl + H_2O + CO_2}$$<br>a) Calculate the number of moles of hydrochloric acid used in the titration.<br>b) Calculate the number of moles of sodium carbonate in the $$25\ \text{cm}^3$$ sample of solution which reacted with the hydrochloric acid.<br>c) How many moles of sodium carbonate were in the $$250\ \text{cm}^3$$ of solution?<br>d) Calculate the mass of sodium carbonate in the $$250\ \text{cm}^3$$ of solution. (Relative atomic masses C = 12, O = 16, Na = 23)<br>e) Using your answer to part (d), what mass of water was in the original powder?<br>f) How many moles of water were present in the original powder? (Relative atomic masses H = 1, O = 16)<br>g) Using your answers to parts (c) and (f), calculate the value of $$x$$.


Correct Answer: 1

Solution

(a) Moles of HCl $$= 0.0256\ \text{dm}^3 \times 0.10 = 2.56 \times 10^{-3}$$ mol.<br>(b) One mole of $$\mathrm{Na_2CO_3}$$ reacts with 2 moles of HCl, so the 25 $$\text{cm}^3$$ sample contained $$1.28 \times 10^{-3}$$ mol.<br>(c) The 250 $$\text{cm}^3$$ is ten times as much, so it contained $$1.28 \times 10^{-2}$$ mol.<br>(d) $$M(\mathrm{Na_2CO_3}) = 2(23) + 12 + 3(16) = 106$$, so the mass is $$0.0128 \times 106 = 1.357\ \text{g}$$.<br>(e) Mass of water $$= 1.59 - 1.357 = 0.233\ \text{g}$$.<br>(f) Moles of water $$= \frac{0.233}{18} = 1.29 \times 10^{-2}$$ mol.<br>(g) $$x = \frac{1.29 \times 10^{-2}}{1.28 \times 10^{-2}} \approx 1$$, so the powder is $$\mathrm{Na_2CO_3 \cdot H_2O}$$.

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