Question

The table below gives the results of three titrations carried out with 0.200 M HCl to determine the molarity of a given NaOH solution using phenolphthalein as indicator. NaOH was taken in the burette and HCl was taken in a conical flask for the titrations

Titration No.	$V_{HCl(mL)}$	$NaOH(mL)$	$M_{NaOH}mol/d{m}^3$
I	21.4	19.3	0.222
II	18.6	16.8	0.221
III	22.2	21.1	0.210

The actual molarity of the prepared NaOH solution was 0.220 mol dm^-3. Which among the following could be the reason for the wrong value obtained in titration III?

• A. Number of drops of phenolphthalein added to the titration flask was more in this titration
• B. The concentration of HCl was wrongly used as 0.250 M for the calculation of M NaOH
• C. A few drops of NaOH solution were spilled outside the titration flask during titration
• D. A few drops of the neutralized solution from titration II were left behind in the flask

Answer 1

Answer: (C)

$\left[O{H}^{-}\right]=\frac{n_{O{H}^{-}}}{V_{O{H}^{-}}}=\frac{n_{acid}}{V_{NaOH}}=\frac{\left[HCl\right]V_{HCl}}{V_{NaOH}}=0.21M<0.220$