rmjracingz6871 rmjracingz6871

25-06-2017
Chemistry

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What volume of 3.3 m hno3 solution is needed to completely neutralize 175 ml of a 4.10 m koh solution?

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dec77cat dec77cat

25-06-2017

HNO₃ + KOH = KNO₃ + H₂O

c(HNO₃)=3.3 mol/L
c(KOH)=4.10 mol/L
v(KOH)=175 mL = 0.175 L

n(KOH)=n(HNO₃)

n(KOH)=c(KOH)v(KOH)
n(HNO₃)=c(HNO₃)v(HNO₃)

c(KOH)v(KOH)=c(HNO₃)v(HNO₃)

v(HNO₃)=c(KOH)v(KOH)/c(HNO₃)

v(HNO₃)=4.10*0.175/3.3=0.217 L = 217 mL

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