A solution of HNO 3 is standardized by reaction with pure sodium carbonate. 2 H + + Na 2 CO 3 ⟶ 2 Na + + H 2 O + CO 2 A volume of 26.66 ± 0.06 mL of HNO 3 solution was required for complete reaction with 0.9479 ± 0.0007 g of Na 2 CO 3 , (FM 105.988 ± 0.001 g/mol ). Find the molarity of the HNO 3 solution and its absolute uncertainty. Note: Significant figures are graded for this problem. To avoid rounding errors, do not round your answers until the very end of your calculations.

= [tex]\frac{(0.8843 + 0.00079 Na_{2}CO_{3})}{(105.988 - 0.001 g/mol} \times \frac{2 mol HNO_{3}}{1 mol Na_{2}CO_{3}} \times \frac{1}{23.72 - 0.06 ml HNO_{3}}[/tex]

= 0.00070584 mol/ml

or, = 0.70584 mol/L (as 1 l = 1000 ml)

The smallest molarity will be calculated as follows.

[tex]\frac{0.8443 - 0.00079 Na_{2}CO_{3}}{105.988 + 0.001 g/mol} \times \frac{2 mol HNO_{3}}{1 mol Na_{2}CO_{3}} \times \frac{1}{(23.72 + 0.06 ml HNO_{3})}[/tex]

= 0.00070115 mol/ml

= 0.70115 mol/l

Now, we will find the average as follows.

[tex]\frac{(0.70584 + 0.70115)}{2}[/tex] = 0.703495

Now, half of the difference will be calculated as follows.

[tex]\frac{(0.70584 - 0.70115)}{2}[/tex]

= 0.002

therefore, we can conclude that the resulting answer is 0.703 ± 0.002 mol/L [tex]HNO_{3}[/tex].