Describe how the molarity of your NaOH solution would be affected if you carried
ID: 493309 • Letter: D
Question
Describe how the molarity of your NaOH solution would be affected if you carried out the titrations to dark pink endpoint. Explain your answer! Describe how the molarity of your NaOH solution would be affected if your burette had an air bubble in the tip when you recorded the initial burette reading, but the bubble was gone when you recorded the final burette reading. Explain your answer! Describe how the molarity of your NaOH solution would be affected if you did not transfer all of the solid KHP into your flask before beginning the titration with NaOH. Explain your answer!Explanation / Answer
1) When the titration is carried out till dark pink endpoint, the volume of the NaOH consumed will be more(as it takes more drops of NaOH from light pink to dark pink) and hence molarity of the NaOH after the calculation will be less than the actual or standard Molarity of NaOH
Explanation: Let say we have 10 ml of the compound with 0.1 M which will be titrated against NaOH. Let's say that 5 ml of the NaOH is consumed for light pink color then the calulation will be,
V1M1 = V2M2
10 x 0.1 = 5 x M2 => M2 = 0.2 M
If we do the titration till the dark pink , it will take few more drops of NaOH, say 8 ml of NaOH. now the molarity of NaOH is 10 x 0.1 = 8 x M2 => M2 = 0.125 M ( which is less than 0.2 M)
2) If the buret tip has air bubble, we don't have NaOH present in that area, so when we measure the amount of NaOH consumed, it will give high reading of the NaOH consumed(the same case as above) , hence the Molarity of the NaOH obtained from the calculation will be lower than the actual Molarity.
3) If we did not transfer the complete KHP in to the flask:
Lets say we have to use 10 ml of KHP with molarity 0.2 M againt the NaOH using a indicator. Lets say the volume of the NaOH consumed is 10 ml. Now the molarity of the NaOH will be
10 x 0.2 = 10 X M2 => M2 = 0.2 M
But actually we have transferred only 8 ml of KHP which consumed 10 ml of NaOH, now the value of M2 will be
8 x 0.2 = 10 x M2 => M2 = 0.16 M
So, if we do not transfer the complete KHP , the Molarity of the NaOH will be shown high (0.2 M) than the actual value (0.16 M)