A solution of HNO3 is standardized by reaction with pure sodium carbonate. 2H++Na2CO3⟶2Na++H2O+CO2 A volume of 23.45±0.05 mL of HNO3 solution was required for complete reaction with 0.9616±0.0009 g of Na2CO3 , (FM 105.988±0.001 g/mol ). Find the molarity of the HNO3 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.
Answers
Answer:
(0,7738±0,0009) M of HNO₃
Explanation:
The reaction of standarization of HNO₃ with Na₂CO₃ is: 2 HNO₃ + Na₂CO₃ ⇒ 2 Na⁺ + H₂O + CO₂ + 2NO₃⁻ To obtain molarity of HNO₃ we need to know both moles and volume of this acid. The volume is (23.45±0.05) mL and to calculate the moles it is necessary to obtain the Na₂CO₃ moles and then convert these to HNO₃ moles, thus: 0,9616 g of Na₂CO₃ × ( 1 mole / 105,988 g) = 9,073×10⁻³ mol Na₂CO₃ × ( 2 moles of HNO₃ / 1 mole of Na₂CO₃) = 1,815×10⁻² moles of HNO₃ Molarity is moles divide liters, thus, molarity of HNO₃ is: 1,815×10⁻² moles / 0,02345 L = 0,7738 M of HNO₃ The absolute uncertainty of multiplication is the sum of relative uncertainty, thus: ΔM = M× (0,0009/0,9616 + 0,001/105,988 + 0,05/23,45) = 0,7738 M× 1,16×10⁻³ = 8,9×10⁻⁴ M Thus, molarity of HNO₃ solution and its absolute uncertainty is: (0,7738±0,0009) M of HNO₃
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