a) How would a too low molarity of NaOH affect the calculated molar mass of the unknown acid? b) How would it affect the calculated Ka of the unknown acid?
If mistakenly calibrated pH meter using pH6 buffer instead of pH7 buffer, all ph meter readings were too low.... how does this affect the molar mass of the unknown acid? would this affect the determined pKa of unknown acid?
3 answers
I need more information as to the nature of your experiment and the procedure used.
Titration experiment to determine the molar mass and dissociation constant of a weak acid, using NaOH
for a). notice that L x M = moles and moles x molar mass = grams; therefore,
L x M x molar mass = grams and
molar mass = g/(L*M) so M too low means molar mass is too high.
b). Did you use the pH meter to record the pH after drop wise addition of NaOH, then plot the data to find the equivalence point? If so, AND you determined the equivalence point from the plotted data on the graph, the molar mass should not be affected. That's because the equivalence point is determined by the slope ON THE GRAPH. If your instructor told you that the equivalence point would occur at such an such a pH and the pH meter read too low, then L NaOH will be too low. Substitute a too low value for L NaOH in the part a formula we used to get the answer. For the pKa value, no matter how the equivalence point is determined, the pH reading will be too low at the point at which Ka is determined so the pKa will be too low.
L x M x molar mass = grams and
molar mass = g/(L*M) so M too low means molar mass is too high.
b). Did you use the pH meter to record the pH after drop wise addition of NaOH, then plot the data to find the equivalence point? If so, AND you determined the equivalence point from the plotted data on the graph, the molar mass should not be affected. That's because the equivalence point is determined by the slope ON THE GRAPH. If your instructor told you that the equivalence point would occur at such an such a pH and the pH meter read too low, then L NaOH will be too low. Substitute a too low value for L NaOH in the part a formula we used to get the answer. For the pKa value, no matter how the equivalence point is determined, the pH reading will be too low at the point at which Ka is determined so the pKa will be too low.
1 answer