A 50 ml sample of calcium hydroxide solution required 19.50 ml of 0.1059 N HCL in a totration. Calculate the following: a. Amount of Ca(OH) in the 50ml sampke b. The Ca(OH)2 concentration in % w/v.

a. 1.05 x 10⁻³ mol

b. 0.155% w/v

Given

50 ml Ca(OH)₂

19.5 ml of 0.1059 N HCl

Required

Amount of Ca(OH)₂

The Ca(OH)₂ concentration in % w/v.

Solution

Titration formula :

M₁V₁n₁=M₂V₂n₂

or

N₁V₁=N₂V₂

n = acid base valence=amount of H⁺/OH⁻(Ca(OH)₂=2, HCl=1)

a.

Input the value(1=Ca(OH)₂, 2= HCl) :

M₂=N₂=0.1059 M

M₁. 50 ml . 2 = 0.1059 . 19.5 . 1

M₁ = 0.021

Amount of Ca(OH)₂ :

mol Ca(OH)₂ = 0.021 x 50 ml = 1.05 mlmol = 1.05 x 10⁻³ mol

b. mass of Ca(OH)₂

= mol x MW

= 1.05 x 10⁻³ mol x 74 g/mol

= 0.0777 g

%w/v = (g solute / volume of solution) x 100

%w/v =( 0.0777 g/ 50 ml) x 100 =0.155% w/v

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the average time it took the car to travel 0.50 meters was 2.61 seconds

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