Question

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?

Answer 1

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

Step-by-step 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) * \frac{\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) * \frac{\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} * \frac{\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.}`