Question

Suppose that 20.0 mL of 0.010 M Ca(OH)2 is required to neutralize 12.0 mL of aqueous HCl solution. What is the molarity of the HCl solution

Answer 1

The molarity of the 12 mL HCl solution needed to neutralise 20 mL 0.010 M Ca(OH)₂ solution is 0.033 M

Balanced equation

2HCl + Ca(OH)₂ —> CaCl₂ + 2H₂O

From the balanced equation above,

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

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

How to determine the molarity of HCl

• Volume of acid, HCl (Va) = 12 mL
• Volume of base, Ca(OH)₂ (Vb) = 20 mL
• Molarity of base, Ca(OH)₂ (Mb) = 0.01 M

• Molarity of acid, HCl (Ma) =?

MaVa / MbVb = nA / nB

(Ma × 12) / (0.01 × 20) = 2

(Ma × 12) / 0.2  = 2

Cross multiply

Ma × 12  = 0.2 × 2

Ma × 12 = 0.4

Divide both side by 12

Ma = 0.4 / 12

Ma = 0.033 M

Thus, the molarity of the HCl solution is 0.033 M

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