Answer to Question #271107 in Chemistry for Vignesh

Question

calculate normality and titre of koh -solution (volume 5.45 ml) if it is necessary for the titration 6.7 ml h2so4-solution with cn=0.1025 mol/l

Answer 1

Solution:

C_N(KOH) × V(KOH) = C_N(H₂SO₄) × V(H₂SO₄)

Therefore,

C_N(KOH) = C_N(H₂SO₄) × V(H₂SO₄) / V(KOH)

C_N(KOH) = (0.1025 mol L^-1 × 6.7 mL) / (5.45 mL) = 0.1260 mol L^-1

Normality of KOH solution is 0.1260 mol L^-1

C_N(KOH) = C_M(KOH)

Molarity of KOH solution is 0.1260 mol L^-1

The molar mass of KOH is 56.1 g mol^-1

Therefore the titre is:

(0.1260 mol L^-1) × (56.1 g mol^-1) = 7.0686 g L^-1 = 7.0686 mg mL^-1

Titre of KOH solution is 7.0686 mg mL^-1

Answer:

Normality of KOH solution is 0.1260 mol L^-1

Titre of KOH solution is 7.0686 mg mL^-1