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

Answer:

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

Explanation:

No. of moles = $\frac{mass}{molar mass}$

= $\frac{0.254}{204}$ = 0.001245 moles

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

= $\frac{0.001245}{0.02}$ = 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.