The correct answer is D according to the answer key Can some

The correct answer is D, according to the answer key. Can someone please explain why?

The table below shows the data for three titrations to determine the concentration of a NaOH solution with standard 0.200 M HCl solution using phenolphthalein as the indicator Trial Vol HCI,Vol NaOH,MNaoH, calc. 21.43 18.57 22.20 19.26 16.73 21.14 0.223 0.222 0.210 Which explanation best accounts for the lower value of the NaOH M in Trial 3? Some of the neutralized solution from Trial 2 was (A) (B) (C) (D) left in the flask for Trial 3 The number of drops of phenolphthalein was doubled in Trial 3 The HCl concentration was used as 0.250 M in the NaOH molarity calculation A few drops of NaOH solution were spilled on the desktop in Trial 3.

Solution

As the concentration of NaOH is trial three is less i.e. the volume of NaOH consumed to neutallise HCl is more.

It has nothing to do with the neutralised solution being left in the flask.as we know NaCl is the neutralised product of this titration. Once NaCl is formed it has no effects on the colour change.

Increasing or Decreasing the amount of indicator will just increase or decrease the intensity of colour change

If the concentration of HCl is incresed to 0.250M .Then according to the formula M1V1=M2V2 NaCl concentration would be 0.2627 M which is not the case

So, only option left is D whic is correct