I think 1 is right.
I disagree with 2. Yes, the solution of NaOH M is too low since all of the NaOH has not dissolved but during the second titration the NaOH is stronger which means it takes less NaOH than it should and M = mmols/mL so smaller NaOH mL means a larger M.
3. The added CO2 reacts with the NaOH to form Na2CO3. If you use methyl red as an indicator it titrates the NaOH there plus all of the Na2CO3 formed so there is no difference.
Titration was used to determine the molarity of acetic acid in vinegar. A primary standard solution of KHP was used to standardize the NaOH.
1. When performing this experiment, impure KHP was used to standardize the NaOH solution. If the impurity is neither acidic or basic, will the percent by mass id acetic acid in the vinegar solution determined by the student be too high or low? Justify answer with explanation.
It would be too high right due to a low number of moles of KHP right?
2. When preparing a NaOH solution, a student did not completely dissolve the NaOH pellets before standarizing the solution with KHP, but by the time the student had refilled the buret with NaOH to titrate the acetic acid, the pellets had dissolved. Will the molarity of acetic acid be too high or low?
Too low right because there is too little of NaOH dissolved leading to low number of moles?
3. How does dissolved CO2 in distilled water affect the accuracy of the determination of a NaOH solution's concentration?
The CO2 in the distilled water will create a more acidic solution which will affct the determination of NaOH solution's concentration? Would it decrease it the concentration of NaOH solution's concentration?
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