What volume of a 0.500 M HCl solution is needed to neutralize each: 14.0 ml of a 0.300 M NaOH and 19.0ml of a 0.200 M Ba (OH) 2

Question

Scarlet Graves

Mathematics

10 January, 10:52

What volume of a 0.500 M HCl solution is needed to neutralize each: 14.0 ml of a 0.300 M NaOH and 19.0ml of a 0.200 M Ba (OH) 2

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Kingston Powers · Answer 1

First thing, we need to know the chemical equation for each neutralization reaction. Then, determine the amount in moles needed hydrochloric acid. We do as follows:

HCl + NaOH = NaCl + H2O

0.300 M NaOH (.014 L) (1 mol HCl / 1 mol NaOH) (1 L / 0.500 M HCl) = 0.0084 L HCl solution or 8.4 mL of 0.500 M HCl

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

0.200 M Ba (OH) 2 (0.019 L) (2 mol HCl / 1 mol Ba (OH) 2) (1 L / 0.500 M HCl) = 0.0152 L HCl solution or 15.2 mL of 0.500 M HCl