Chem...involving titrations

First, calculate the no. of moles of the remaining HCl
HCl + NaOH --> NaCl + H2O
no of moles of NaOH: 0.4 x (15/1000) = 0.006mol
since the mole ratio of NaOH : HCl is 1:1
so, no. of moles of the remaining HCl = 0.006mol

Secondly, calculate the original HCl
2HCl + Na2CO3 --> 2NaCl + H2O + CO2
no. of moles of the original HCl: 0.5 x (35/1000) = 0.0175mol

Next, calculate the no. of moles of HCl used
0.0175 - 0.006 = 0.0115mol

Then calculate the no. of moles of Na2CO2 reacted
since the mole ratio of HCl : Na2CO3 is 2:1
so, the no. of moles of Na2CO3 reacted is 0.0175/2 = 0.00575mol

Then calculate the mass of reacted Na2CO3
0.00575 x (23+23+12+48) = 0.6095g

Purity:
0.6095/0.7 x 100% = 87.07%

35cm³ of 0.500M HCl were added to 0.70g of the sample of Na2CO3 containing some NaCl as impurity. The excess acid was found to be neutralised by 15.00 cm³ of 0.400M NaOH solution. Calculate the percentage purity of the Na2CO3 in the sample.


Consider the neutralisation of excess HCl:
HCl + NaOH → NaCl + H2O
Molar ratio HCl : NaOH = 1 : 1
No. of moles of NaOH used = 0.400 x (15/1000) = 0.00600 mol
No. of moles of excess HCl = 0.00600 mol

Consider the reaction between HCl and Na2CO3:
2HCl + Na2CO3 → 2NaCl + H2O + CO2
Mole ratio HCl : Na2CO3 = 2 : 1
Total no. of moles of HCl = 0.500 x (35/1000) = 0.0175 mol
No. of moles of HCl reacted with Na2CO3 = 0.0175 - 0.000600 = 0.0115 mol
No. of moles of Na2CO3 = 0.0115 x (1/2) = 0.00575 mol
Molar mass of Na2CO3 = 23x2 + 12 + 16x3 = 106 g/mol
Mass of Na2CO3 in the sample = 0.00575 x 106 = 6.10 g
% purity of the Na2CO3 in the sample = (6.10/7.00) x 100% = 87.1%