Question

Q2.  0.254 g of KHP (204 g/mol) titrated against 20.0 mL of unknown NaOH (40.0 g/mol) solution to get the end point of phenolphthalein indicator. Calculate the concentration of NaOH in g/L unit.​

Answer 1

Mass concentration (g/L) = 2.49g/L.

Step-by-step explanation:

No. of moles =
`(mass)/(molar mass)`

=
`(0.254)/(204)` = 0.001245 moles

Concentration of KHP (C1) in litres = n/v

=
`(0.001245)/(0.02)` = 0.062 mol/L

We know that:

`C_(1) V_(1)` =
`C_(2) V_(2)`

where c1v1 and c2v2 are the products of concentration and volumes of KHP and NaOH respectively.

Since mole ratio is 1 : 1.

1 mole of NaOH - 40g

0.001245 mole of NaOH = 40 × 0.001245 = 0.0498g

⇒0.0498g of NaOH was used during the titration

∴Mass concentration (g/L) = 0.0498g ÷ 0.02L

= 2.49g/L.