Question

wha mass of Ca(OH)2 is required to neutralize 50.00mL of 0.180 M HCl?

Answer 1

Task:

what mass of $\mathrm{Ca(OH)_2}$ is required to neutralize $50.00\mathrm{mL}$ of 0.180 M HCl?

Solution:

The chemical equation for this reaction is

$$\mathrm{Ca(OH)_2} + 2\mathrm{HCl} = \mathrm{CaCl_2} + 2\mathrm{H_2O}$$

The formula for molar concentration is

$$\mathrm{C(M)} = \mathrm{n(mol)} / \mathrm{V(L)}$$

n- amount of the substance (mol)

V – volume of solution (L)

That’s why

$$\mathrm{n(mol)} = \mathrm{C(M)} \cdot \mathrm{V(L)}$$

From the chemical equation

$$\mathrm{n(Ca(OH)_2)} = \mathrm{n(HCl)} / 2 = \mathrm{C(HCl)} \cdot \mathrm{V(HCl)} / 2$$ $$\mathrm{n(Ca(OH)_2)} = 0.180 \cdot 50.00 \cdot 10^{-3} / 2 = 4.50 \cdot 10^{-3} \mathrm{mol}$$

The mass of $\mathrm{Ca(OH)_2}$ is

$$\mathrm{m(Ca(OH)_2)} = \mathrm{n(Ca(OH)_2)} \cdot \mathrm{MW(Ca(OH)_2)}$$

From the periodic table of elements $\mathrm{MW(Ca(OH)_2)} = \mathrm{MW(Ca)} + 2\mathrm{MW(O)} + 2\mathrm{MW(H)} = 40.0 + 2 \cdot 16 + 2 \cdot 1.0 = 74\ \mathrm{g/mol}$

$$\mathrm{m(Ca(OH)_2)} = 4.50 \cdot 10^{-3} \cdot 74 = 0.33\ \mathrm{g}$$

Answer: $\mathrm{m(Ca(OH)_2)} = 0.33\ \mathrm{g}$