Question

1. Calculate the molar amount of KHP used to neutralize the NaOH solution.

2. Calculate the molar concentration of the NaOH solution that you prepared.

3. Compare the actual molarity of your NaOH solution with your goal of 0.10 M.

Equivalance point-25.3,

Measurements found in lab.

Mass of NaOH needed=0.40g

Mass of KHP needed=0.505g

Answer 1

Rewrite the problem in a more intelligible way:

0.505 g of Potassium Hydrogen Phthalate (KHP) was neutralized with 25.3 mL of NaOH solution 0.10 M (nominal concentration).

KHP Molar mass = 204.22 g/mol. KHP is pure and air-stable, so it is used as primary standard for acid-base titrations (as mono-protic acid).

NaOH Molar mass = 39.9971 g/mol (monobase), aproximate here as 40.00

KHP      + NaOH     ?      KNaP      +      H₂0

1 mol         1 mol

204.22 g     40.00 g

1. The KHP quantity is

0.505/204.22 = 0.002473 mol

2. The same NaOH quantity was neutralized, i.e. 0.002473 mol.

C_NaOH = quantity/volume used for titration = 0.002473 mol/ 0.0235 L = 0.10523 M

3. For comparison, calculate a correction factor f for the nominal concentration of NaOH:

            f = actual concentration/ nominal concentration

            f = 0.10523/0.10000 = 1.0523   (dimenssionless)

Note: The mass m 0.40 g NaOH is the calculated quantity of NaOH needed to prepare 100 mL solution 0.1M NaOH:

m = concentration x volume x molar mass = 0.1 mol/L x 0.100 L x 40 g/mol =

    = 0.400 g

Or Volume = m / (conc. x molar mass) = 0.400 / (0.1 x 40) = 0.100 L