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9 May, 13:40

How many milliliters of 0.100 M Ba (OH) 2 are required to neutralize 20.0 mL of 0.250 M HCl?

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  1. 9 May, 13:53
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    Ba (OH) ₂ + 2HCl = BaCl₂ + 2H₂O

    n (HCl) = c (HCl) v (HCl)

    n{Ba (OH) ₂) }=c{Ba (OH) ₂}v{Ba (OH) ₂}=2n (HCl) = 2c (HCl) v (HCl)

    v{Ba (OH) ₂}=2c (HCl) v (HCl) / c{Ba (OH) ₂}

    v{Ba (OH) ₂}=2*0.250*0.020/0.100=0.100 L = 100 mL
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