Chemistry Question

Question

A solution of F− is prepared by dissolving 0.0805±0.0006 g NaF (molar mass = 41.989±0.001 g/mol) in 120.00±0.06 mL of water. Calculate the concentration of F− in solution and its absolute uncertainty.

Answer 1

[F⁻] = 0.01598 ± 0.00012 M

Step 1 — the value

mol NaF = 0.0805 g ÷ 41.989 g/mol = 1.9172 × 10⁻³ mol

NaF dissociates 1:1, so mol F⁻ = mol NaF.

[F⁻] = 1.9172 × 10⁻³ mol ÷ 0.12000 L = 0.015976 M

Step 2 — which rule applies

Every operation here is a multiplication or a division. For those, you propagate relative uncertainties in quadrature — not absolute ones. (Absolute uncertainties add in quadrature only for addition and subtraction. Mixing the two rules up is the most common error in this whole topic.)

Step 3 — relative uncertainties

mass: 0.0006 / 0.0805 = 0.00745  (0.745%)
molar mass: 0.001 / 41.989 = 0.0000238  (0.00238%)
volume: 0.06 / 120.00 = 0.000500  (0.0500%)

Step 4 — combine

e_rel = √(0.00745² + 0.0000238² + 0.000500²)

e_rel = √(5.555 × 10⁻⁵ + 5.7 × 10⁻¹⁰ + 2.50 × 10⁻⁷) = √(5.580 × 10⁻⁵) = 0.00747  (0.747%)

Step 5 — back to absolute

e_abs = 0.015976 M × 0.00747 = 1.19 × 10⁻⁴ M

[F⁻] = 0.01598 ± 0.00012 M  (about 0.0160 ± 0.0001 M)

The thing actually worth taking away from this problem

Look at what each term contributed. The mass uncertainty is 0.745%; everything else is at most 0.05%. Squaring makes that gap enormous — the molar mass term is 5.7 × 10⁻¹⁰ against the mass term's 5.6 × 10⁻⁵, five orders of magnitude smaller. Drop the molar mass and volume terms entirely and you still get 0.745% instead of 0.747%.

That is how quadrature behaves in general: the largest single term dominates, and anything under about a third of it is invisible. Practically, it means a more precise balance is the only thing that would meaningfully improve this measurement. Buying a better volumetric flask would do essentially nothing. Being able to see that at a glance is the real skill this question is training.

One formatting note: report the uncertainty to one significant figure (0.0001) or at most two (0.00012), and round the value to match that decimal place. Writing 0.0159764 ± 0.00012 claims precision the last digits do not have.