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Kobotan [32]
3 years ago
13

How many milliliters of a 0.285 M HCl solution are needed to neutralize 249 mL of a 0.0443 M Ba(OH)2 solution?

Chemistry
1 answer:
meriva3 years ago
8 0

Answer:

\large \boxed{\text{77.4 mL}}

Explanation:

                Ba(OH)₂ + 2HCl ⟶ BaCl₂ + H₂O

    V/mL:     249

c/mol·L⁻¹:  0.0443     0.285

1. Calculate the moles of Ba(OH)₂

\text{Moles of Ba(OH)$_{2}$} = \text{0.249 L Ba(OH)}_{2} \times \dfrac{\text{0.0443 mol Ba(OH)}_{2}}{\text{1 L Ba(OH)$_{2}$}} = \text{0.011 03 mol Ba(OH)}_{2}

2. Calculate the moles of HCl

The molar ratio is 2 mol HCl:1 mol Ba(OH)₂

\text{Moles of HCl} = \text{0.011 03 mol Ba(OH)}_{2} \times \dfrac{\text{2 mol HCl}}{\text{1  mol Ba(OH)}_{2}} = \text{0.022 06 mol HCl}

3. Calculate the volume of HCl

V_{\text{HCl}} = \text{0.022 06 mol HCl} \times \dfrac{\text{1 L HCl}}{\text{0.285 mol HCl}} = \text{0.0774 L HCl} = \textbf{77.4 mL HCl}\\\\\text{You must add $\large \boxed{\textbf{77.4 mL}}$ of HCl.}

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