If the stockroom fails to dry the KHP to remove the waters of hydration, the student's calculated concentration
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)
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.
To determine the effect on the calculated concentration of the standardized base, let's consider each scenario separately:
1. If the stockroom fails to dry the KHP to remove the waters of hydration:
The presence of water molecules in the KHP can affect the accuracy of the titration. This is because KHP is a solid acid that contains water molecules within its crystal structure. If the stockroom fails to remove these water molecules, they will contribute to the weight of the KHP sample, resulting in a higher apparent concentration of KHP. This, in turn, would lead to a lower calculated concentration of the standardized base.
Therefore, if the stockroom fails to dry the KHP properly, the calculated concentration for the standardized base will be **low**.
2. A drop of NaOH falls into the KHP solution before the titration:
The addition of a drop of NaOH before the titration begins will increase the volume of the base solution in the reaction. This increased volume will dilute the concentration of the base. Consequently, when the student performs the titration, the apparent concentration of the standardized base will appear lower than its actual concentration. However, the true concentration of the base remains unaffected by the presence of one drop of NaOH.
Therefore, if a drop of NaOH falls into the KHP solution before the titration, the calculated concentration for the standardized base will be **unaffected**.
If the stockroom fails to dry the KHP to remove the waters of hydration, the student's calculated concentration for the standardized base will be unaffected.
If a drop of NaOH is hanging from the buret tip before the titration of KHP and then falls into the KHP solution, the calculated concentration for the standardized base will be affected. It will be high.