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
I hope it helps!
