a) "Na_2CO_3 (aq) + 2HCl(aq) = 2NaCl (aq) + CO2 (g) + H_2O (l)"
"C_1V_1 =C_2V_2"
As mole ratio "n(Na_2CO_3):n(HCl ) = 1:2"
then "C_1V_1 =\\frac{C_2V_2}{2}"
"C_1 = ?"
"V_1 = 25 cm^3"
"C_2 = 0.101 M"
"V_2 = 24.35 cm^3"
"C_1 = 0.0492 M"
b) We have 1000 cm3 of Na2CO3 solution with concentration of 0.0492 M. Then, there are 0.0492 moles of Na2CO3 in this solution. We are said in the task that firstly we had 25 ml of saturated solution and then we diluted it with water till 1000 cm3. Then all of the 0.0492 moles firsty were in 25 ml of saturated solution. As this solution (i.e. 25 ml) was saturated one, find the solubility of Na2CO3:
"solubility=\\frac{0.0492 mol}{0.025 dm^3} = 1.97 \\frac{mol}{dm^3}"
c ) As solubility of Na2CO3 for 1000 cm3 solution is 1.97 mol/dm3, then when all water will be evaporated, 1.97 mol of Na2CO3 will be left. Find the mass of Na2CO3:
"1.97 mol Na_2CO_3 (\\frac{106 gNa_2CO_3}{1 mol Na_2CO_3}) = 209 g"
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