I. A student made the following mistakes while performing the titration experime
ID: 695612 • Letter: I
Question
I. A student made the following mistakes while performing the titration experiment. Explain how it will affect the results of the experiment. i. The student forgot to add the indicator. ii. The initial volume of NaOH solution in the buret was not 0.00 mL, but 5.12 ml iii. The student forgot to add 40 mL water into the flasks 1 and 2. 2. A standard solution of 1.25 M HCI was used to determine the concentration of 52.8 mL of an unknown solution of potassium hydroxide (KOH). If 12.6 mL of the acid was required to reach endpoint, determine the molarity of the KOH solution? 3. What volume of a 0.5310 M solution of HNO, would be needed to titrate 120.00 ml. of a 0.1525 M solution of NaOH?Explanation / Answer
Ans. #1. A. The indicator is required to determine the endpoint of titration. So, without the indicator, the endpoint can’t be determined. As a result, the volume of standard NaOH required to reach the endpoint would also be undetermined. In turn, the concertation of unknown acid can’t be determined, too because the concetration of acid is determined using amount of NaOH consumed to reach th endpoint.
# B. If the initial volume of NaOH is 5.12 mL instead of 0.0 mL, the apparent volume of NaOH consumed would be lesser than the actual volume.
At the titration endpoint, C1V1 (NaOH) = C2V2 (acid). That is the volume of NaOH is prop rational to concertation (C2) of acid.
Therefore, lower volume of NaOH consumed would lead to lower than actual concertation of acid after calculation.
#C. At the titration endpoint, the number of moles of NaOH consumed is equal to the number of moles of acid in the sample. Since addition of water or dilution does not affect the number of moles of acid (though it lowers the concertation), the calculated concentration of acid remains unaffected.
#2. Balanced reaction: HCl(aq) + KOH(aq) -----------> KCl(aq) + H2O(l)
In the balanced reaction, 1 mol HCl is neutralized by 1 mol KOH.
So, at the titration endpoint, C1V1 (KOH) = C2V2 (HCl).
Or, C1 x 12.6 mL = 1.25 M x 52.8 mL
Or, C1 = (1.25 M x 52.8 mL) / 12.6 mL
Hence, C1 = 5.238 M
Therefore, required concentration of KOH = 5.238 M
#3. Balanced reaction: HNO3(aq) + NaOH(aq) -----------> KNO3(aq) + H2O(l)
In the balanced reaction, 1 mol HNO3 is neutralized by 1 mol NaOH.
So, at the titration endpoint, C1V1 (NaOH) = C2V2 (HNO3).
Or, 0.1525 M x 120.0 mL = 0.5310 M x V2
Or, V2 = (0.1525 M x 120.0 mL) / 0.5310 M
Hence, V2 = 34.463 mL
Therefore, required volume of HNO3 = 34.463 M