How to Calculate Absolute Error in Moles for an Acid-Base Titration Experiment?

[thinktank75]

How do you calculate the absolute error of an experiment (in moles of acides) when you are given:

18.59 +/-0.02mL of a 12.85 +/ 0.03M NaOh solution that is going to be neutralized with an unknown acid?

I'm not sure how to plug it into the equation, or I may be using the wrong equation.
Thanks

 

[thinktank75]

I took the greatest possible error (18.59 + 12.85) then subtracted it by the maximum error (18.59 + 0.02) + (12.85 + 0.03) but i keep getting the wrong answer, am I suppose to convert the mL to M? (they're both different, but I can't find the mL for NaOh since I don't have the moles) ...

 

[Blueshift5]

To calculate the absolute error in this experiment, you will need to use the formula for absolute error, which is the absolute value of the difference between the measured value and the true value. In this case, the measured value is 18.59 +/- 0.02 mL and the true value is 12.85 +/- 0.03 M.

First, calculate the absolute error in volume by subtracting the lower bound (18.59 - 0.02 = 18.57) from the upper bound (18.59 + 0.02 = 18.61). This gives you a range of 18.57 - 18.61 mL for the volume. Since we are looking for the absolute value, we can take the average of these two values, which is 18.59 mL.

Next, calculate the absolute error in concentration by subtracting the lower bound (12.85 - 0.03 = 12.82) from the upper bound (12.85 + 0.03 = 12.88). This gives you a range of 12.82 - 12.88 M for the concentration. Again, taking the average of these two values gives you 12.85 M.

To calculate the absolute error in moles of acid, you will need to use the formula for molarity (M = moles of solute / liters of solution). Rearranging this equation, we get moles of solute = M x liters of solution. Therefore, the absolute error in moles of acid will be the absolute value of the difference between the upper and lower bounds for the moles of NaOH, which is 12.85 M x 18.59 mL - 12.85 M x 18.61 mL = 0.03 moles.

In summary, the absolute error in moles of acid for this experiment is 0.03 moles. It is important to note that this calculation assumes that the uncertainties in volume and concentration are independent and do not affect each other. If this is not the case, a more complex method such as propagation of uncertainty may need to be used to calculate the absolute error.