Question

Calculate the molar concentration of the acid if "50.6 mL" of hydrochloric acid was required to neutralize 45.0 mL of 0.345 M "calcium hydroxide".

Answer 1

The molar concentration of the acid, HCl required for the reaction is 0.614 M

We'll begin by writing the balanced equation for the reaction. This is given below:

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

The mole ratio of the acid, HCl (nA) = 2

The mole ratio of the base, Ca(OH)₂ (nB) = 1

• From the question given above, the following data were obtained:

Volume of acid, HCl (Va) = 50.6 mL

Volume of base, Ca(OH)₂ (Vb) = 45 mL

Molarity of base, Ca(OH)₂ (Mb) = 0.345 M

Molarity of acid, HCl (Ma) =?

MaVa / MbVb = nA/nB

(Ma × 50.6) / (0.345 × 45) = 2

(Ma × 50.6) / 15.525 = 2

Cross multiply

Ma × 50.6 = 15.525 × 2

Ma × 50.6 = 31.05

Divide both side by 50.6

Ma = 31.05 / 50.6

Ma = 0.614 M

Therefore, the molar concentration of the acid, HCl is 0.614 M

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Answer 2

The molar concentration of the acid is 0.614 mol/L

The balanced chemical reaction for the reaction is

Ca(OH)₂ + 2HCl → CaCl₂ + H₂O

Using CV/C'V' = n/n' where C = molar concentration of hydrochloric acid, V = volume of hydrochloric acid = 50.6 mL = 0.0506 L, n = number of moles of hydrochloric acid = 2, C' = molar concentration of calcium hydroxide = 0.345 M = 0.345 mol/L, V' = volume of calcium hydroxide = 45.0 mL = 0.045 L, n' = number of moles of calcium hydroxide = 1.

Making C subject of the formula, we have

C = nC'V'/n'V

Substituting the values of the variables into the equation, we have

C = nC'V'/n'V

C = 2 × 0.345 mol/L × 0.045 L/(1 × 0.0506 L)

C = 2 × 0.345 mol/L × 0.045 L/0.0506 L

C = 0.03105 mol/0.0506 L

C = 0.614 mol/L

So, the molar concentration of the acid is 0.614 mol/L

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