If the stockroom fails to dry the KHP to remove the waters of hydration, the student's calculated concentration

Question

for the standardized base will be

(low, high, or unaffected)
A drop of NaOH is hanging from the buret tip before the titration of KHP. Once the student begins the titration that drop falls into the KHP solution. The calculated concentration for the standardized base wil be
(low, high, or unaffected)

Answer 1

I'll do the first one so you can see how it's done. First, write the formulas used to determine the molarity of the base.
moles KHP = grams KHP/molar mass KHP
moles KHP = moles NaOH.
moles NaOH = mLNaOH x M NaOH and
MNaOH = moles NaOH/mL NaOH.

If the KHP is not dry, grams in equation 1 will be too high (because you are weighing water + KHP) and that means moles KHP will be high (technically, moles KHP will be lower BUT we think there is more there so moles will be too high).
Move to equation 2. moles KHP too high gives moles NaOH too high.
Move to equation 3. Moles NaOH too high gives M too high.
(You CAN go about this in reverse fashion but some students get mixed up this way. Choose the method best for you.
If we weigh out wet KHP, I get too little KHP (because some of it is water).
Move to equation 2. Too little KHP gives too few moles KHP and that means it will less volume NaOH to titrate the fewer moles. Move to equation 3. Since
M NaOH = molesNaOH/mL NaOH, too few mL in the denominator gives too large M NaOH.)
Finally, I think it easier to do it this way. Put all of the equations together.
mL NaOH x M NaOH x molar mass KPH = g KHP and solve for M NaOH.

M NaOH = g KHP/mL NaOH x molar mass KHP.
Then if grams are too high, M NaOH is too high. Or the reverse way; if KHP is contaminated with water, too little KHP means too little volume in the titration and fewer mL means M too high.