chemistry
posted by Fai on .
A 0.4671g sample containing sodium bicarbonate was titirated with hcl requiring 40.72ml. The acid was standardized by titrating 0.1876g of sodium carbonate FW 106mg/mmol requiring 37.86ml of the acid. Find the percentage of nahc03 fw 84mg/mmol in the sample.
My own calculation:
meq hcl = meq na2co3
Na2co3 =106/2
= 53
37.86x=187.6mg/53
37.86x=3.5
x=3.5/37.86
x=0.0935
0.0935 x 40.72 x 84/46.71 x 100 = 68.4%
Who check for me the answer 68.4%

Na2co3 =106/2
= 53
37.86x=187.6mg/53
37.86x=3.5
x=3.5/37.86
x=0.0935
0.0935 x 40.72 x 84/46.71 x 100 = 68.4%
If you multiply this out you get 684.46; the error is you should have divided by 467.1. Then I would not truncate the 68.46% but leave it to four significant figures.
Who check for me the answer 68.4%