Question

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

Answer 1

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`