The answers depends upon what you are doing and you don't provide enough information; however, here is my best guess as to what is going on.
You must be titrating an acid with a base(NaOH). The mols base used = M x L and you know both M and L. You are using the pH meter to know when you've reached the equivalence point.
Here are the two equations you're using.
equation 1 is M x L = moles NaOH = moles acid.
equation 2 is molar mass = grams/moles.
Now we subject these two equations to the problems.
a. pH meter reads lower (meaning it reads more acidic than the solution actually is) so you must add more base to get to the equivalence. More mL base = a higher number from equation 1 and if you plug a high number of mols into equation 2, that means a lower molar mass is calculated.
b. NaOH missed the mark means more must be added to reach equivalence point, more mL from equation means more moles to equation 2 and smaller molar mass.
c. I don't know what you did in c. If you weighed out a solid acid, placed ALL of it into the beaker, then added 75 instead of 50 mL, the molar mass will be affected because you've titrated all of the sample weighed. If, however, you weighed out the solid acid, placed it in the beaker, added 75 instead of 50 mL, then transferred a PORTION of that to titrate, then you transferred less than intended because of the dilution, that means fewer mL to reach the equivalence point, moles from equation 1 will be a smaller number and a smaller value for moles in equation 2 means a larger molar mass.
Hope this helps.
How would the calculated value of the molar mass of an unknown acid be affected (higher, lower, or no change) if the following occurs?
(a) The pH meter was incorrectly calibrated to read lower than the actual pH.
(b) During the titration, several drops of NaOH missed the reaction beaker and fell onto the bench top.
(c) Acid was dissolved in 75 mL of distilled water rather than 50 mL.
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