Consider the reaction of KOH and HNO2.
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Consider the reaction of KOH and HNO2.

[From: ] [author: ] [Date: 12-12-08] [Hit: ]
290 = 0.0.......
Calculate the milliliters of 0.290 M KOH solution required to neutralize 36.0 mL of a 0.250 M HNO2 solution.

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First you need to get in moles to make the transition. Recall that Molarity * Liters = Moles
So... x*0.290M = 0.036L * 0.250M
in essence we are saying here that the moles of KOH = the moles of HNO_2_
using algebra solve for x
(0.036*0.250) / 0.290 = 0.031 L
0.031 *1000 = 31 mL of KOH required
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