Question

5.00 mL of 0.10 M oxalic acid solution taken in a conical flask is titrated against NaOH from a burette using phenolphthalein indicator. The volume of NaOH required for the appearance of permanent faint pink color is tabulated below for five experiments. What is the concentration, in molarity, of the NaOH solution?

Answer 1

Answer:

50 is the answer of your questions

Answer 2

we have to find the concentration in molarity of the NaOH solution..

solution : we know, no of equivalents = (weight in gm)/(equivalent weight)

= (weight in gm)/(Molecular weight/n - factor)

= no of moles × n - factor

at neutralisation,

no of equivalents of oxalic acid = no of equivalents of NaOH

⇒no of moles of oxalic acid × n - factor = no of moles of NaOH × n - factor

⇒ 0.1 M × 5ml/1000 × 2 = molarity of NaOH × 9ml/1000 × 1 [ here we take volume of NaOH is 9ml because successive the last three experiments give same date i.e., 9 ml so the volume of NaOH would be 9 ml ]

⇒1 = molarity of NaOH × 9

⇒molarity of NaOH = 1/9 = 0.11 M

Therefore the molarity of NaOH solution is 0.11 M