Question

Titration and standardization

50mL buret is filled with 50mL of 1M NaOH solution (6.02g of NaOH in 150mL water). 1.55g of KHP is completely dissolved in water. Then titration of NaOH solution is done into KHP solution.

1) How do you standardize NaOH with KHP data?

2) How do you determine the volume of NaOH required to neutralize the KHP solution?

3) How do you determine the molarity of the sodium hydroxide solution?

Answer 1

Given:-

wt.of NaOH = 6.02 g

volume of water used to make NaOH solution = 150 ml = 150 /1000 = 0.15 L

wt.of potassium hydrogen phthalate (KHP) = 1.55 g

As we know that

molar mass of NaOH = molar mass of Na + molar mass of O + molar mass of H

molar mass of NaOH = 23 + 16 + 1

molar mass of NaOH = 40 g / mol

(3)-

As we know that

molarity of compound = wt. of compound (g) / molar mass of compound( g / mol) volume of solution in liter(L)

molarity of NaOH = wt. of NaOH(g) / molar mass of NaOH( g / mol) volume of solution in liter(L)

molarity of NaOH = 6.02 g / 40 g / mol 0.15 L

molarity of NaOH = 6.02 mol / 40    0.15 L

molarity of NaOH = 6.02 mol / 6.0 L

molarity of NaOH = 1.0 mol / L or 1 M (i.e the answer)

suppose 1.55 KHP is dissolved in 100 ml of water to make the solution of KHP then

wt.of potassium hydrogen phthalate (KHP) = 1.55 g

molr mass of  potassium hydrogen phthalate (KHP) = 204.2 g / mol

volume of solution = 100 ml = 0.1 L

molarity of KHP= wt. of KHP(g) / molar mass of KHP( g / mol) volume of solution in liter(L)

molarity of KHP= 1.55 g / 204.2 g / mol   0.1 L

molarity of KHP = 1.55 mol / 20.42 L

molarity of KHP = 0.0759 mol/ L

molarity of KHP = 0.0759 M

(1)- Standardization of  NaOH with KHP is means that to find out the required volume of 1.0 M of NaOH filled in the buret to completely neutralized the 0.0759 M KHP solution taken 100 ml in volumetric flask i.e volume of KHP solution is 100 ml. This standardization of NaOH confirmed the molarity of the NaOH solution experimentally.

(2)- As we know that

molarity of KHP (M₁) = 0.0759 M

voulme of KHP (V₁) = 100.0 ml

molarity of NaOH (M₂) = 1.0 M

volume of NaOH (V₂) = ?

As we know that according to the formula

M₁V₁ = M₂V₂

0.0759 M 100.0 ml =  1.0 M V₂

V₂ =0.0759 M 100.0 ml / 1.0 M

V₂ = 0.0759 100.0 ml

V₂ = 7.59 ml ( i.e volume of NaOH )

The answer is 7.59 ml of NaOH required to neutralize the 1.55 g KHP dissolved in 100 ml water solution. It can be done by titrating 1.0 M NaOH taken in buret against 100 ml of 0.0759 M of KHP solution.