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?
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.
1. If impure KHP is used to standardize the NaOH solution, the percent by mass of acetic acid in the vinegar solution determined by the student will be too low. This is because the impurities present in the KHP will cause the student to measure a lower volume of NaOH solution required to neutralize the acetic acid in the vinegar. As a result, the calculated molarity of the acetic acid will be higher than the actual value.
2. If the NaOH pellets were not completely dissolved before standardizing the solution with KHP, the molarity of acetic acid in the vinegar solution determined by the student will be too high. This is because the incomplete dissolution of the NaOH pellets will result in a lower concentration of NaOH in the solution used for titration. As a result, more NaOH will be required to neutralize the acetic acid in the vinegar, leading to a higher calculated molarity of the acetic acid.
3. The dissolved CO2 in distilled water can affect the accuracy of determining the NaOH solution's concentration. This is because CO2 can react with water to form carbonic acid (H2CO3), which is a weak acid. The presence of carbonic acid in the solution can lead to a higher apparent concentration of H+ ions, which can interfere with the titration process. As a result, the determined concentration of the NaOH solution may be higher than its actual concentration.
1. The percent by mass of acetic acid in the vinegar solution determined by the student will be too high when impure KHP is used to standardize the NaOH solution. This is because if the impurity in the KHP is neither acidic nor basic, it will not react with the NaOH solution during the standardization process. As a result, the student will obtain a higher value for the molarity of the NaOH solution, leading to a higher calculated molarity for the acetic acid in the vinegar.
2. If the student did not completely dissolve the NaOH pellets before standardizing the solution with KHP, but the pellets had dissolved by the time the student titrated the acetic acid, the molarity of the acetic acid will be too high. This is because the incomplete dissolution of NaOH initially would have led to a lower concentration of NaOH in the solution, resulting in a higher calculated molarity for the acetic acid during the titration.
3. Dissolved CO2 in distilled water can affect the accuracy of the determination of a NaOH solution's concentration. When CO2 dissolves in water, it forms carbonic acid (H2CO3), a weak acid. This increases the acidity of the solution and results in protonation of some water molecules, generating H3O+ ions. The presence of the H3O+ ions can react with NaOH during the titration, leading to neutralization reactions and the consumption of NaOH. This would result in a falsely lower calculated concentration of the NaOH solution. Therefore, the accuracy of the determination of the NaOH solution's concentration can be negatively affected by dissolved CO2, leading to a decrease in the calculated concentration.