Question

A 25.00 mL volume of an HA acid solution of unknown concentration with 2 drops of phenolphthalein requires 0.1320 M NaOH was put in buret to reach the endpoint in the titration. The initial reading & fina reading of NaOH is shown in the picture Initial 9.63 mL 24.16 ml Volume of HA acid (mL) Volume of NaOH (mL) Molar concentration of NaOH (molL)

Answer 1

Answer – We are given, Volume of HA acid = 25.00 mL

Initial volume of NaOH = 9.63 mL

Final volume of NaOH = 24.16 mL

Volume of NaOH = 24.16-9.63 = 14.53 mL

[NaOH] = 0.1320 M

Now, moles of NaOH dispensed –

Moles of NaOH = 0.1320 M x 0.01453 L

                             = 0.00192 moles

Now, the molar concentration of HA acid –

We know the reaction HA and NaOH is as follow –

HA + NaOH ----> NaA + H₂O

mole ratio is 1:1

so, moles of NaOH = moles of HA = 0.00192 moles

we know molar concentration = moles / volume (L)

Molar concentration of HA acid = 0.00192 moles / 0.025 L

                                                         = 0.0767 M