You have a bottle labeled “53.4 (±0.4) % w/w NaOH” with a density of 1.52 (±0.01) g/mL. You use 16.7 (± 0.2) mL of the NaOH to prepare 2.00 (±0.02) L of 0.169M NaOH. Report the absolute error for the 0.169M of NaOH. You may assume negligible error in the NaOH molar mass.

Question

You have a bottle labeled “53.4 (±0.4) % w/w NaOH” with a density of 1.52 (±0.01) g/mL. You use 16.7 (± 0.2) mL of the NaOH to prepare 2.00 (±0.02) L of 0.169M NaOH. Report the absolute error for the 0.169M of NaOH. You may assume negligible error in the NaOH molar mass.

Answer 1

http://web.uvic.ca/~jalexndr/192UncertRules.pdf

Answer 2

To calculate the absolute error for the 0.169M NaOH solution, we need to consider the errors in the measurements of the concentration, volume, and density.

Let's start by calculating the absolute error in the mass of NaOH used. We know the volume used is 16.7 mL, and the density of the NaOH is 1.52 g/mL. Therefore, the mass of NaOH used can be calculated as follows:

Mass = Volume x Density
Mass = 16.7 mL x 1.52 g/mL
Mass = 25.384 g

The absolute error in the mass can be calculated by taking into account the error in the volume and error in the density. The maximum error in the volume is given as ±0.2 mL, and the maximum error in the density is given as ±0.01 g/mL. To calculate the absolute error, we multiply the errors by the corresponding values used in the equation:

Absolute Error in Mass = Volume Error x Density + Volume x Density Error
Absolute Error in Mass = (0.2 mL x 1.52 g/mL) + (16.7 mL x 0.01 g/mL)
Absolute Error in Mass = 0.304 g + 0.167 g
Absolute Error in Mass = 0.471 g

Next, we need to calculate the moles of NaOH used. The molar mass of NaOH is approximately 40 g/mol. To find the moles, we divide the mass by the molar mass:

Moles = Mass / Molar Mass
Moles = 25.384 g / 40 g/mol
Moles = 0.6346 mol

Now, we can calculate the absolute error in the moles of NaOH used by multiplying the absolute error in mass with the molar mass:

Absolute Error in Moles = Absolute Error in Mass x (1 / Molar Mass)
Absolute Error in Moles = 0.471 g x (1 / 40 g/mol)
Absolute Error in Moles = 0.011775 mol

Finally, we can calculate the absolute error in the molarity of the NaOH solution. We know the moles used and the volume of the solution prepared (2.00 L). The formula to calculate molarity is:

Molarity = Moles / Volume

Absolute Error in Molarity = Absolute Error in Moles / Volume
Absolute Error in Molarity = 0.011775 mol / 2.00 L
Absolute Error in Molarity = 0.0058875 M

Therefore, the absolute error in the 0.169M NaOH solution is ±0.0058875 M.