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

Answer:

v = 500 milliliters

Explanation:

$HNO_{3}$  ⇒  $H^{+} + NO^{3-}$

$KOH$  ⇒  $K^{+} + OH^{-}$

1 $H^{+}$ to 1 $OH^{-}$

$\frac{0,5}{V} = \frac{0,1}{100} \\\\v * 0,1 = 50\\v = 500 milliliters$