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Chapter 25: Problem 1

An accurately weighed amount \((3.6284 \mathrm{~g}\) ) of hydrated sodium carbonate \(\left(\mathrm{Na}_{2} \mathrm{CO}_{3} \times \mathrm{H}_{2} \mathrm{O}\right)\) was dissolved in water \((250.00 \mathrm{~mL}\) ). An aliquot \((25.00 \mathrm{~mL})\) of this solution required hydrochloric acid solution \((25.4 \mathrm{~mL} ; 0.100 \mathrm{M})\) for equivalence. Calculate the number of molecules of water of crystallisation \((x)\) in the original hydrated sodium carbonate. (HINT: you need to calculate \(M_{\mathrm{r}}\) for the original \(\left.\mathrm{Na}_{2} \mathrm{CO}_{3} \times \mathrm{H}_{2} \mathrm{O} .\right)\)

Short Answer

Expert verified
The number of molecules of water of crystallisation 'x' in the original hydrated sodium carbonate (\(Na_2CO_3 \cdot xH_2O\)) is approximately 10.

Step by step solution

01

Calculate moles of hydrochloric acid

First, calculate the number of moles of hydrochloric acid (HCl) used for equivalence by using the molarity formula, Molarity = moles/volume (in liters). Here, the provided molarity \(M = 0.100 \mathrm{M}\) and volume \(V = 25.4 \, \mathrm{mL} = 0.0254 \, \mathrm{L}\). The number of moles \(n_{\mathrm{HCl}}\) is then given by \(n_{\mathrm{HCl}} = M \times V = 0.100 \, \mathrm{M} \times 0.0254 \, \mathrm{L} = 0.00254 \, \mathrm{mol}\).
02

Determine moles of Sodium Carbonate

In the balanced equation for the reaction between sodium carbonate (\(Na_2CO_3\)) and hydrochloric acid (\(HCl \rightarrow 2NaCl + CO_2 + H_2O\)), each one mole of sodium carbonate reacts with two moles of hydrochloric acid. So, the moles of sodimum carbonate in the 25.00 mL aliquot \(n_{Na_2CO_3} = 0.00254 \, \mathrm{mol}/2 = 0.00127 \, \mathrm{mol}\).
03

Calculate total moles of Sodium Carbonate in solution

Since 25.00mL of the sodium carbonate solution contains 0.00127 mol of sodium carbonate, the total amount of sodium carbonate in solution can be calculated proportionally. For a total volume of 250.00mL, \(N_{Na_2CO_3} = 0.00127 \, \mathrm{mol} \times (250.00 \, \mathrm{mL}/25.00 \, \mathrm{mL}) = 0.0127 \, \mathrm{mol}\).
04

Calculate molar mass (Mr) of Sodium Carbonate

The molar mass \(M_{Na_2CO_3}\) is given by the molecular weight of Sodium Carbonate, \(\_M_{Na_2CO_3} = 2 \times M_{Na} + M_{C} + 3 \times M_{O} = 2 \times 23 + 12 + 3 \times 16 = 106 g/mol\).
05

Calculate molar mass (Mr) of Hydrated Sodium Carbonate

Given that 3.6284 g of hydrated sodium carbonate forms 0.0127 mole, then the molar mass (\(M_{Hydrated Na_2CO_3}\)) of hydrated sodium carbonate can be entirely calculated by \(M_{Hydrated Na_2CO_3} = \mathrm{Mass/Mole} = 3.6284 g/0.0127 mol = 285.7 g/mol\).
06

Determine number of water molecules in Hydrated Sodium Carbonate formula

Finally, to find 'x', the total molar mass of the hydrated sodium carbonate is equal to the molar mass of sodium carbonate plus the mass of 'x' water molecules. The molar mass of water \(_M_{H_2O} = 2 \times M_{H} + M_{O} = 2 \times 1 + 16 = 18 g/mol\). Therefore using the equation \(M_{Hydrated Na_2CO_3} = M_{Na_2CO_3} + x \times {M_{H_2O}}\), plug in the molar masses and solve for 'x' to get \(x = (M_{Hydrated Na_2CO_3} - M_{Na_2CO_3}) / {M_{H_2O}} = (285.7 g/mol - 106 g/mol) / 18 g/mol = 9.98 \approx 10\).

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