What volume of 0.130M HNO3 will react completely with 44.0mL of 0.250M NaOH?
Answer:
HNO3 + NaOH ---> NaNO3 + H2O
since there is no need to balance you'll have this equation
MV (HNO3) 1 divided by
=
MV(NaOH) 1
0.130M *V (HNO3) = (44.00/1000) * 0.250M
V(HNO3) = 0.0846L (3 sig fig) or 84.6 mL
mL acid x Normality of Acid = mL Base x Normality of Base
44 x .25 / .13 = 84.6 mL
normality is the same as Molarity for Nitric Acid and Sodium Hydroxide
balanced equation
HNO3 + NaOH -> NaNO3 + H20
so the realtionship is 1 mol to 1 mol
44 ml of .25 M NaOH has .25 * .044 (.011 mol) of NaOH
so you need and equal amount of HNO3
.130 M = .011 mol / X L
x = .0846 L of .130 M HNO3
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