Question

A CHEM 2242 student dissolved 69.322 g of an unknown monoprotic acid in a 250 mL volumetric flask. The student then transferred 5.00 mL of this solution into a conical flask and diluted the 5.00 mL with another 85.00 mL of DI H₂O. Using the second derivative method the student estimated that equivalence occured when 36.86 mL of 0.1066 M NaOH titrant had been added. What is the molarity of the unknown acid solution?

Answer 1

Answer – We are given, mass of unknown monoprotic acid = 69.322 g

Volume of flask = 250 mL out of this 250 mL student take out 5 mL and diluted with another 85 mL water, so the final volume is 5+85 = 90 mL

When this is reacted with 36.86 mL of 0.1066 M NaOH equivalence occurred

So first we need to calculate moles of NaOH

Moles of NaOH = 0.1066 M * 0.03686 L

                           = 0.003929 moles

We know the given acid is monoprotic acid , so we assume it is HA

So reaction is –

HA + NaOH ------> NaA + H₂O

So 1 moles of NaOH = 1 moles of HA

So, 0.00392 moles of NaaOH = ?

= 0.003929 moles of HA

So molarity of HA = 0.003929 moles / 0.100 L

                                = 0.03929 M

So this is the after diluted,

So, original molarity of acid, M1 = ? , V1 = 5 mL , V2 = 100 mL , M2 = 0.03929 M

So by diluting law

M1V1 = M2V2

M1 = M2V2/V1

      = 0.03929 M * 100 mL / 5 mL

      = 0.786 M

So the molarity of the unknown acid solution is 0.786 M