## And I can't calculate to obtain that answer either. I don't think the problem is worded correctly. Here is why I think that.

If the sample is 56.7% NaHCO3 then it is 43.3% Na2CO3.

g Na2CO3 = 2 x 0.433 = 0.866g Na2CO3

g NaHCO3 = 2 x 0.567 = 1.134g NaHCO3

total = 1.134+0.866 = 2.000g.

We titrated 0.1 of that so we titrated

0.0866g Na2CO3 + 0.1134g NaHCO3 and that is 0.0866+0.1134 = 0.2000 g titrated. How many mL of 0.087N HCl SHOULD THIS TAKE?

mL x N x mew = grams or

mL = g/(N*mew)

For Na2CO3 mL = 0.0866/(0.087*0.053) = 18.78 mL to titrate the 0.0866g Na2CO3.

For NaHCO3 mL = 0.1134/(0.087*0.084) = 15.52 mL to titrate the 0.1134g NaHCO3

.

mL for Na2CO3 + mL for NaHCO3 = total mL = 18.78+15.52 = 34.3 mL and for the problem that should be 22.5 mL. It isn't; therefore, the problem is not possible to work as is. If the problem is changed so that the HCl used was 34.3 mL of 0.087N then it is worked as follows:

eqn 1 is X + Y = 2

eqn 2 is [0.1X/0.053]+ [0.1Y/0.084] = 34.3*0.087

From eqn 2 we have

1.8867X + 1.19Y = 2.984

From eqn 1 if X+Y = 2 than X = 2-Y. Substitute that into eqn 2 as follows.

1.8867(2-Y) + 1.19Y = 2.984

3.7734 - 1.8867Y + 1.19Y = 2.984

-0.6976Y = -0.7893

Y = 1.133g NaHCO3

% = (1.133/2)*100 = 56.6% NaHCO3