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Answer to Question #169855 in General Chemistry for Alexis

Question #169855

How many mL of .145 of M HCL are needed to completely neutralize 49.0 mL of 0.100 M Ba(OH)2 solution


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Expert's answer
2021-03-08T07:09:43-0500

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


n(Ba(OH)2) = c * V = 0.1M * 0.049L = 0.0049mol

n(HCl) = 2n(Ba(OH)2) = 0.0098mol

V(HCl) = n / c = 0.0098mol / 0.145M = 0.0676L = 67.6mL

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