Btw, KHP is Potassium hydrogen phthalate
C8H5KO4
What would the effect on your molarity of NaOH be if you had added 35 mL of water to dissolve the KHP (acid) instead of 25 mL? Explain.
My answer so far is that adding more water shouldn't affect the molarity at all because water does not account for the reaction between NaOH and KHP.
This explanation seems rather simple, am I missing something or am I completely wrong? Thanks in advance!
5 answers
Molarity is another way of saying molar concentration. If you know the formula for molarity you will see that the change in volume will affect the molarity of a solution. The more water you add the more you will dilute the solution.
Yes I know the formula of molarity M=mol/L but the volume of water is not accounted for in this calculation specifically since the volume we're using is the volume of NaOH solution dispensed into the flask. Therefore, water shouldn't have an effect on the molarity of the NaOH.
I think the question MAY (that's MAY) not be worded properly. Two scenarios.
1. IF the question is stated properly, then the molarity of the KHP solution created by added 35 mL water instead of 25 mL water is decreased. As Mia notes, M = mols/L and if L is larger than M is smaller. BUT, I don't think that's the question because the question is about the M of NaOH and not M KHP.
2. I think the question REALLY is that of a titration of NaOH vs KHP. If in that titration, one adds 35 mL water to the KHP solid, adds NaOH and titrates the KHP solution formed, then the calculated value for M NaOH is the same no matter what volume of water is added to the KHP before titrated with NaOH. The reason is that the KHP, as weighed, contains mols KHP = grams/molar mass KHP. That will take so many mols NaOH and that determines the M of the NaOH. The volume of water added prior to titration doesn't matter. The indicator in the titration changes when mols KHP = mol NaOH. Water not entering into the reaction is part of a good explanation of what is going on but the crucial answer is that mols KHP = mols NaOH and the amount of water added is that required for the convenience of the operator.
1. IF the question is stated properly, then the molarity of the KHP solution created by added 35 mL water instead of 25 mL water is decreased. As Mia notes, M = mols/L and if L is larger than M is smaller. BUT, I don't think that's the question because the question is about the M of NaOH and not M KHP.
2. I think the question REALLY is that of a titration of NaOH vs KHP. If in that titration, one adds 35 mL water to the KHP solid, adds NaOH and titrates the KHP solution formed, then the calculated value for M NaOH is the same no matter what volume of water is added to the KHP before titrated with NaOH. The reason is that the KHP, as weighed, contains mols KHP = grams/molar mass KHP. That will take so many mols NaOH and that determines the M of the NaOH. The volume of water added prior to titration doesn't matter. The indicator in the titration changes when mols KHP = mol NaOH. Water not entering into the reaction is part of a good explanation of what is going on but the crucial answer is that mols KHP = mols NaOH and the amount of water added is that required for the convenience of the operator.
DrBob222 you're exactly right on 2! Sorry my teacher didn't really clarify exactly what was being asked but I took on the question like how #2 is stated, and therefore said that the molarity of the NaOH wouldn't be affected by the amount of water added prior to the reaction.
0 answers