[tongleecheen]

If I were preforming a weak acid-strong base titration with 0.1 M NaOH and 0.2 g of an unknown acid mixed with 50 mL of water, how could I explain the following errors if I thought the values the nominal values but they were actually off? Explain errors on Ka and molar mass. The experimental pKa is 4.7 and the molar mass is 58 g/mol.
a)   If the concentration of NaOH was actually greater than the nominal value provided (0.10 M).
b)   If the mass of acid was actually greater than the nominal value provided (0.20 g). 
c)   If the volume of the acidic solution was actually greater than the nominal value (50.0 mL).  Based on these answers, provide two reasons why your experimental values for the molar mass and pKa differ from the literature values. 

[Borek]

You have to show your attempts at solving the problem to receive help, this is a forum policy, but I will give you a small hint:

If the concentration of NaOH was actually greater than the nominal value provided (0.10 M).

Try to do calculations for the nominal concentration and for 0.11 M. Compare results.

[tongleecheen]

Oh sorry,
for a without calculations I said that the Ka would end up being larger because you will end up using less NaOH to titrate the acid, therefore the half equivalence point would also be shifted toward the left. I am not super sure if there is a shift in the change in pH because Ka is supposed to be constant right? And that Ka only changes with a temperature change. I said the molar mass would increase because less NaOH needed would result in a lower moles concentration at equivalence so if you divide 0.2 g by a small number, it would result in a larger mm.
For b I said no change in Ka because the Ka ratio would adjust for any differences and that the mm would decrease because you need more NaOH, so you will be dividing 0.2 g by a larger number of moles.
For C, I said an increase in Ka because it is a more dilute solution, so you will need less NaOH to reach the equivalence. Basically the same reasoning as part a. Less NaOH results in a left shifted pKa and that results in a larger Ka. The mm would increase because you are dividing 0.2 g of acid by a smaller mole amount.

[Borek]

Lets start from the beginning: what were you asked to determine and how?

[tongleecheen]

Explain the effect of each of the following systematic errors on the experimentally determined values of Ka and the molar mass compared to the literature values. 
a)   If the concentration of NaOH was actually greater than the nominal value provided (0.10 M).
b)   If the mass of acid was actually greater than the nominal value provided (0.20 g). 
c)   If the volume of the acidic solution was actually greater than the nominal value (50.0 mL).  Based on these answers, provide two reasons why your experimental values for the molar mass and pKa differ from the literature values. 
A weak acid-strong base titration was preformed. It was determined that weak acid was acetic acid with a molar mass of 59 g/mol. The equivalence point was reached when 34 mL of 0.1 M NaOH was added. 0.2 grams of acid was mixed with enough solvent to make a 50 mL solution.

[Borek]

How do you determine Ka?

[tongleecheen]

Ka was determined from the half equivalence point, where pH=pKa. The half equivalence point was at 17 mL of NaOH added to the acid which resulted in a pH of 4.67. Exponentiating we got a Ka of roughly 2.138E-5.

[Borek]

Does your calculation of half equivalence point depend on the concentration of NaOH?