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Molarity and absolute uncertainty

We can measure the concentration of HCl solution by reaction with pure sodium carbonate: 2H+ + Na2CO3 --> 2Na+ + H2O + CO2. Complete reaction with 0.967 4 (+/-0.000 9) g of Na2CO3 (FM 105.988 4 (+/- 0.000 7)) required 27.35 (+/- 0.04) mL of HCl. Find the molarity of the HCl and its absolute uncertainty.

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Answer #1
0.9674(+/-0.0009)g Na2CO3 * (1 mol / 105.9884(+/-0.0007)g) = ?


e%= sqr[(0.0009 / 0.9674 *100)^2 + (0.0007 / 105.9884 *100)^2]

e%= 0.093034%


0.9674 / 105.9884= 0.009127(+/-0.093034)mol Na2CO3


0.009127 * 0.093034= +/-0.0008


0.009127(+/-0.0008)mol Na2CO3 * (2mol HCl / 1mol Na2CO3)= 0.018254(+/-0.0016)mol HCl


27.35(+/-0.04)mL HCl * (1L/1000mL)= 0.02735(+/-0.00004)L HCl


M=mol/L

0.018254(+/-0.0016)mol HCl / 0.02735(+/-0.00004)L HCl = ?


e%= sqr[(0.0016)^2 + (0.00004/0.02735 *100)^2]

e%= 0.146%


0.018254 / 0.02735= 0.667


(0.667 * 0.146) / 100= +/-0.001


The molarity of HCl is 0.667(+/-0.001)M
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