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What volume of 0.500 M HNO₃(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?
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What volume of 0.500 M HNO₃(aq) must completely react to neutralize 100.0 milliliters of 0.100 M KOH(aq)?

User RTF
by
6.8k points

1 Answer

1 vote

Answer:

v = 500 milliliters

Step-by-step explanation:


HNO_(3) ⇒
H^(+) + NO^(3-)


KOH ⇒
K^(+) + OH^(-)

1
H^(+) to 1
OH^(-)


(0,5)/(V) = (0,1)/(100) \\\\v * 0,1 = 50\\v = 500 milliliters

User Mike Glukhov
by
6.2k points
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