Question

4. If you expect the unknown acid molarities to be between 0.010 M and 0.20 M and the sodium hydroxide solution to be 0.050 M, what volume of unknown acid solution would you place into the flask? State your reasoning or assumptions that you make (perhaps typical of those given in question 3, perhaps not) and show your calculation.

Bonus. State two possible errors when using burets that would reduce the accuracy of your titration. Explain specifically whether the result would be a calculated acid molarity that is smaller than the actual value or a calculated acid molarity that is larger than the actual value.

Answer 1

4) I do not have the exact question; seems like you missed to put in the complete information. Additionally, it seems that the acid that you are titrating will have a molarity between 0.1 M and 0.2 M. The molarity values that you have provided for the acid, i.e, 0.01 M to 0.2 M is an extremely large range and we cannot use 0.05 M sodium hydroxide to titrate 0.01 M acid. A concentrated base cannot be used to titrate a dilute acid.

I will assume that you meant the acid molarity to be between 0.1 M and 0.2 M and that you delivered the sodium hydroxide via a pipette and the acid was added from the burette.

Since pipettes have fixed capacities like 10, 20 or 25 mL, you can deliver a fixed concentration of sodium hydroxide. Let us consider you used a 25 mL pipette and the molarity of sodium hydroxide = 0.050 M.

Moles of sodium hydroxide delivered = (volume of NaOH delivered in L)*(molarity of NaOH) = (25 mL)*(1 L/1000 mL)*(0.050 mole/L) = 0.00125 mole.

Further, we will assume that the acid-base reaction takes place on a 1:1 molar ratio (you haven’t provided the exact data and hence we are working with assumptions).

Therefore, moles of acid neutralized = moles of NaOH added = 0.00125.

Let us first assume the molarity of the acid to be 0.1 M; therefore, the volume of acid added from the burette = moles of acid neutralized/molarity of acid = (0.00125 mole)/(0.1 mole/L) = 0.0125 L = (0.0125 L)*(1000 mL/1 L) = 12.5 mL.

Now, if the molarity of the acid is 0.2 M, then the volume of acid added = (0.00125 mole)/(0.2 mole/L) = 0.00625 L = 6.25 mL.

Therefore, if the molarity of the acid added is between 0.1 and 0.2 M, then a volume of acid between 6.25-12.5 mL will be required for the titration (ans).

Sources of error:

i) Air is trapped in the stop-cock or at the tip of the burette. Such an error will lead to delivery of a lower volume of the acid and hence the calculated acid molarity will be higher than the actual value. Additionally, if the solution in the burette contains air bubbles, then also a lower volume of acid is delivered leading to a higher calculated molarity.

ii) The solution from the burette is added too rapidly without proper stirring of the titration flask. This leads to a local build-up of the acid and the solution in the flask is not properly homogenized. Therefore, a higher volume of the acid will need to be delivered from the burette to bring about the end point and the calculated molarity of the acid will be lower than the actual value.