How many milliliters of a .5M HNO3 (aq) is needed to completely neutralize 100mL of a 0.10M KOH (aq)?
volumeacid*concentration acid in N= volume base*concentration base inN
Va*.5*1=100ml*.1*1
Va=200ml
To determine the number of milliliters of a 0.5M HNO3 (aq) needed to completely neutralize 100mL of a 0.10M KOH (aq), you can use the stoichiometry of the balanced chemical equation.
The balanced chemical equation for the neutralization reaction between HNO3 and KOH is as follows:
HNO3(aq) + KOH(aq) → KNO3(aq) + H2O(l)
From the equation, you can see that the stoichiometric ratio between HNO3 and KOH is 1:1. This means that one mole of HNO3 reacts with one mole of KOH.
First, you need to calculate the number of moles of KOH in 100mL of the 0.10M KOH solution:
moles of KOH = concentration of KOH × volume of KOH solution
= 0.10 M × 0.100 L (converting 100mL to liters)
= 0.010 moles
Since the stoichiometric ratio between HNO3 and KOH is 1:1, you need an equal number of moles of HNO3 to neutralize the KOH.
Next, you can calculate the volume of the 0.5M HNO3 solution required:
volume of HNO3 solution = moles of HNO3 / concentration of HNO3
= 0.010 moles / 0.5 M
= 0.020 L (converting liters to milliliters)
= 20 mL
Therefore, you will need 20 milliliters of the 0.5M HNO3 solution to completely neutralize 100 milliliters of the 0.10M KOH solution.