The true percentage of Na
2CO
3 in a sample of soda ash is 40.24%. A 0.4134 g sample was analyzed by titration with HCl. The sample required 28.56 mL of 0.1106 M HCl for complete neutralization. The equivalent weight of Na
2CO
3 is 53.00. The relative error, in ppt (parts per thousand) would be
A) 13 ppt
B) 6.5 ppt
C) 6.4 ppt
D)12.8 ppt
I started it by doing 0.02856 L*0.1106 M HCl = 0.003313 mol HCl required to titrate the Na
2CO
3. And then I wrote the neutralization equation: Na
2CO
3 + 2HCl
2NaCl + CO
2 + H
2O.
So there's 2 moles of HCl per mole of Na
2CO
3. So I divided the 0.003313 mol HCl by 2 to get 0.00166 mol Na
2CO
3. And then I did 0.4024 * 0.4134 g = 0.1664 g Na
2CO
3. 0.1664 g/0.00166 mol = 105.328 g/mol Na
2CO
3. They gave that the equivalent weight of Na
2CO
3 is 53.00, but that didn't give a reasonable answer, so I redid it and got that the molar mass of Na
2CO
3 is 105.99. and then I did (105.99-105.328)/105.99 * 100 = 6.24 ppt, which is close to B and C, but neither one of those. I realize that it could've been a rounding error, but if there was one, it would be closer to C) 6.4, whereas the answer is B) 6.5. So is there something I did wrong, or is it just like a really huge rounding error or something?