Answer to Question #162452 in Chemistry for Ahmed

Question

Calculate normality and titre of KOH -solution (volume 6.5 ml) if it is 

nessesary for the titration 5.0 ml HCl – solution with CN = 0.1065 mol/l.

Answer 1

Solution:

The balanced chemical equation:

KOH + HCl → KCl + H₂O

According to the equation above:

n(KOH) = n(HCl)

or

C_N(KOH) × V(KOH) = C_N(HCl) × V(HCl)

Hence,

C_N(KOH) = C_N(HCl) × V(HCl) / V(KOH)

C_N(KOH) = 0.1065 mol L^-1 × 5.0 mL / 6.5 mL = 0.08192 mol L^-1

C_N(KOH) = 0.08192 mol/L

T(KOH) = C_N(KOH) × M(KOH) / 1000

The molar mass of KOH is 56.1056 g mol^-1.

Hence,

T(KOH) = (0.08192 mol L^-1 × 56.1056 g mol^-1) / 1000 = 0.004596 g mL^-1

T(KOH) = 0.004596 g/mL

Answer:

Normality of KOH solution is 0.08192 mol/L.

Titre of KOH solution is 0.004596 g/mL.