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14 October, 23:52

6. What volume of 6M Ba (OH) 2 will be needed to neutralize 400.0 ml of a 2.0 M HCl solution?

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  1. 15 October, 00:51
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    133 mL

    Explanation:

    M = mol solute / L solution

    mol solute = M * L solution

    mol solute = (2.0 M * 0.4 L)

    mol solute = 0.8 mol HCl = mol H

    We need the same amount of OH moles to neutralize the solution

    L solution = mol solute/M

    L solution = 0.8 mol Ba (OH) 2 / 6M

    L solution=0.133 L=133 mL
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