Question

How many grams of Ca(OH)₂ are there if a sample is titrated to the complete endpoint with 38.0 mL of 0.120N HCl?

Answer 1

for HCl, normality and molarity is same since HCl has only 1 H+

we have the Balanced chemical equation as:

2 HCl + Ca(OH)2 ---> CaCl2 + 2 H2O

lets calculate the mol of HCl

volume , V = 38.0 mL

= 3.8*10^-2 L

we have below equation to be used:

number of mol,

n = Molarity * Volume

= 0.12*0.038

= 4.56*10^-3 mol

From balanced chemical reaction, we see that

when 2 mol of HCl reacts, 1 mol of Ca(OH)2 reacts

mol of Ca(OH)2 reacted = (1/2)* moles of HCl

= (1/2)*4.56*10^-3

= 2.28*10^-3 mol

This is number of moles of Ca(OH)2

Molar mass of Ca(OH)2 = 1*MM(Ca) + 2*MM(O) + 2*MM(H)

= 1*40.08 + 2*16.0 + 2*1.008

= 74.096 g/mol

we have below equation to be used:

mass of Ca(OH)2,

m = number of mol * molar mass

= 2.28*10^-3 mol * 74.096 g/mol

= 0.169 g

Answer: 0.169 g