Calculate the volume of 2.00 M solution of KOH that is required to neutralize 30.0 mL of 1.00 M solution of HNO3.
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To calculate the volume of the KOH solution required to neutralize the HNO3 solution, we can use the concept of stoichiometry and the equation for the neutralization reaction between KOH and HNO3.
The balanced neutralization reaction between KOH and HNO3 is:
KOH + HNO3 -> KNO3 + H2O
From the balanced equation, we can see that the stoichiometric ratio between KOH and HNO3 is 1:1. This means that every one mole of KOH reacts with one mole of HNO3.
First, let's determine the number of moles of HNO3 in the given 30.0 mL of 1.00 M solution. To do this, we'll use the equation:
moles = concentration (M) × volume (L)
Given:
Concentration of HNO3 = 1.00 M
Volume of HNO3 solution = 30.0 mL = 30.0 / 1000 L = 0.0300 L
moles of HNO3 = 1.00 M × 0.0300 L = 0.0300 moles
Since the stoichiometric ratio is 1:1, the number of moles of KOH required to neutralize the HNO3 is also 0.0300 moles.
Now, let's find the volume (in liters) of the 2.00 M KOH solution required.
Given:
Concentration of KOH = 2.00 M
Using the equation:
moles = concentration (M) × volume (L)
0.0300 moles = 2.00 M × Volume (L)
Volume (L) = 0.0300 moles / 2.00 M = 0.0150 L = 15.0 mL
Therefore, the volume of the 2.00 M solution of KOH required to neutralize 30.0 mL of 1.00 M HNO3 solution is 15.0 mL.