Question

"A" contains 0.101 moldm-3 HCI "B" is obtained by dilluting 25.0cm3 of a saturated solution of anhydrous Na2CO3 at 25C to 1000cm3 with distiled water. 25.00cm3 portions of B requried an average of 24.35cm3 of A for complete neutralization using methylorange as the indicator.
From the above data, calculate the
a. Concentration of solution B in moldm-3
b. Solubility of Na2CO3 at 25C on moldm-3
Mass of Na2CO3 obtained when 1dm3 of the saturated solution was evaporated to dryness 2HCI(aq) + Na2CO3(aq) 2NaCI + H2O + CO2 [Na=23; O=16; C=12]

Answer 1

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 $C_1V_1 =\frac{C_2V_2}{2}$

$C_1 = ?$

$V_1 = 25 cm^3$

$C_2 = 0.101 M$

$V_2 = 24.35 cm^3$

$$C_1\times 25 =\frac{0.101\times 24.35}{2}$$

$$C_1 = 0.0492 M$$

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:

$1.97 mol Na_2CO_3 (\frac{106 gNa_2CO_3}{1 mol Na_2CO_3}) = 209 g$