a) A student fails to wash the weighing paper when transferring the KHP sample into the beaker. What effect does this error have on the calculated molarity of the NAOH solution? Mathematically justify your answer.
b) A student failed to notice an air bubble trapped in the tip of the buret during their experiment. What effect does this error have on the calculated molarity of the NaOH solution. Mathematically justify your answer.
Thank you so much. Also, would this same procedure be used for when there's an air bubble in the titration?
Yes. Just go through and determine how each step is affected, take that answer and move to the next step, then continue until you end up with the last step.
a) Ah, the classic weighing paper mishap! Well, failing to wash the weighing paper actually adds extra mass to the KHP sample being transferred into the beaker. This will cause the calculated molarity of the NaOH solution to be higher than it should be. Mathematically speaking, since molarity is calculated using the formula Molarity = moles/volume, the extra mass of the KHP sample will lead to an overestimation of the moles of KHP. Consequently, the calculated molarity of the NaOH solution will be higher than the actual value. Oopsie daisy!
b) Ah, the sneaky air bubble! If there is an air bubble trapped in the tip of the buret during the experiment, it can affect the volume of NaOH solution that is dispensed. This means that the calculated volume of NaOH solution used will be less than the actual volume, leading to a lower calculated molarity of the NaOH solution. Mathematically, since molarity is calculated using the formula Molarity = moles/volume, a lower volume will result in a higher calculated molarity. Wait, did I say higher earlier? My bad! I meant a lower calculated molarity. So yes, the air bubble will make the calculated molarity lower than it should be. Whoopsie daisy again!
a) When a student fails to wash the weighing paper when transferring the KHP (potassium hydrogen phthalate) sample into the beaker, the residual KHP left on the paper can affect the calculated molarity of the NaOH (sodium hydroxide) solution.
To justify this mathematically, let's consider the balanced chemical equation between KHP and NaOH:
KHP + NaOH -> NaKP + H2O
In this equation, the mole ratio between KHP and NaOH is 1:1. This means that one mole of KHP reacts with one mole of NaOH to form one mole of NaKP (sodium potassium phthalate) and one mole of water.
Now, let's assume that the student did not wash the weighing paper and transferred some amount of KHP along with it. This additional mass of KHP will result in an excess amount of KHP in the reaction.
According to the stoichiometry of the balanced equation, this excess KHP will react with an equal amount of NaOH. As a result, some of the NaOH solution will be consumed in this reaction instead of reacting with the KHP in the beaker.
Since the volume of the NaOH solution is fixed, the actual amount of NaOH available for reaction will be less, leading to a lower calculated molarity.
Mathematically, we can represent this as:
M1V1 = M2V2
Where:
M1 = initial molarity of the NaOH solution
V1 = initial volume of the NaOH solution
M2 = final molarity of the NaOH solution (calculated)
V2 = final volume of the NaOH solution (same as V1)
Due to the excess KHP reacting with NaOH from the weighing paper, the actual volume of NaOH available for reaction will be less than V1. Therefore, the calculated molarity (M2) will be lower than the initial molarity (M1).
b) When a student fails to notice an air bubble trapped in the tip of the buret during their experiment, it can affect the accuracy of the measured volume of NaOH solution delivered. This, in turn, can impact the calculated molarity of the NaOH solution.
To justify this mathematically, let's consider the volume measurement equation:
V = (Vb - Va)
Where:
V = volume of NaOH solution delivered
Vb = final buret reading
Va = initial buret reading
If there is an air bubble trapped in the tip of the buret, it will displace some volume of the NaOH solution. As a result, the measured volume of NaOH solution (V) will be less than the actual volume that was delivered.
Mathematically, we can represent this as:
V_actual = V + ΔV
Where:
V_actual = actual volume of NaOH solution delivered (equals the mole ratio)
ΔV = volume displaced by the air bubble
Since the calculated molarity depends on accurately measuring the volume of NaOH solution delivered, an error in the volume measurement will result in an error in the calculated molarity.
As a result, the calculated molarity of the NaOH solution will be higher than the actual molarity. This is because the measured volume (V) in the calculation will be less than the actual volume of NaOH solution delivered (V_actual), leading to an overestimate of the molarity.
I think the best way to answer this type question is to write the steps involved in the procedure. Here is the KHP/NaOH problem.
1. mols KHP = grams/molar mass
2. mols NaOH = mols KHP
3. M NaOH = mols NaOH/L NaOH
I presume not washing the paper on which the KHP was weighed means some of the KHP was not transferred to the titration flask. That means mass KHP is low. From step 1 that means mols KHP is low. Step 2 says mols NaOH is low. Step 3 says M is low.
(For what's it's worth, I NEVER used weighing paper; as a result I never had that problem.)