If the pH probe was miscalibrated and reports the pH to be 1 unit higher than the true pH of the solution it is in, would the molar mass you report for the solid acid be higher, lower, or equal to the true molar mass of the acid?
This for my post lab write up.
I think that the answer would be higher. I am not sure how or why it would be this. Any help is greatly appreciated.
5 answers
I don't know what lab you did. Did you place a weak acid or base in solution, measure the pH, calculate what? what else do you know?
DrBob222,
In the lab we used a NaOH solution with a 0.1088M to calculate the equivalence point of a unknown weak acid using "Logger Pro" program. I used 0.4530g of the unknown acid and dissolved it in 75ml of distilled water. For my 3 three trials, I gather enough data/information to calculate 1.) the mole of the acid in the sample, 2.) Molar Concentration of the acid, 3.) Average Molar Concentration of acid, 4.) Deviation, 4.) the average deviation, 5.) Ka of the acid, 6.) average Ka of acid.
I titrated the NaOH slowly into the weak acid solution, and collected data point the computer system until I had a decent titration curve. From there I was able to determine the equivalence point.
Does this help?
Michelle
In the lab we used a NaOH solution with a 0.1088M to calculate the equivalence point of a unknown weak acid using "Logger Pro" program. I used 0.4530g of the unknown acid and dissolved it in 75ml of distilled water. For my 3 three trials, I gather enough data/information to calculate 1.) the mole of the acid in the sample, 2.) Molar Concentration of the acid, 3.) Average Molar Concentration of acid, 4.) Deviation, 4.) the average deviation, 5.) Ka of the acid, 6.) average Ka of acid.
I titrated the NaOH slowly into the weak acid solution, and collected data point the computer system until I had a decent titration curve. From there I was able to determine the equivalence point.
Does this help?
Michelle
I am still working on a handful of the calculations.
Sorry to get back so late but here is what I think based on what you wrote.
IF (and only IF) you used the titration CURVE to determine the equivalence point on the computer (that is you determined the eq pt graphically on the computer), then mols acid comes from mols NaOH and that is mols = M NaOH x L NaOH. The M doesn't change and the L NaOH is determined by the graph and NOT from the reading of pH, the molar mass = grams/mols. grams is always the same, mols is the same so the molar mass should not change. The Ka will change and pKa value will be too high by 1 unit, that is, a pKa of 3 would read as pKa 4. None of what I wrote above will be true if you titrated to the pH value for the salt of the weak acid. In that case the L NaOH will be affected (too small since you think you are at the equivalence point but you aren't) then molar mass = grams/mols and molar mass will be too large.
IF (and only IF) you used the titration CURVE to determine the equivalence point on the computer (that is you determined the eq pt graphically on the computer), then mols acid comes from mols NaOH and that is mols = M NaOH x L NaOH. The M doesn't change and the L NaOH is determined by the graph and NOT from the reading of pH, the molar mass = grams/mols. grams is always the same, mols is the same so the molar mass should not change. The Ka will change and pKa value will be too high by 1 unit, that is, a pKa of 3 would read as pKa 4. None of what I wrote above will be true if you titrated to the pH value for the salt of the weak acid. In that case the L NaOH will be affected (too small since you think you are at the equivalence point but you aren't) then molar mass = grams/mols and molar mass will be too large.
Ok. Thank you. That is what I was thinking. I really appreciate your help.
1 answer