A 0.500 M HNO3 solution was used to titrate a 21.15 mL KOH solution. The endpoint was reached after 22.30 mL of titrant were delivered. Find the molar concentration of KOH in the original solution.
To solve for the molar concentration of KOH in the original solution, we can use the neutralization equation and the formula for moles and molar concentration.
Neutralization Equation:
HNO3 + KOH → KNO3 + H2O
Moles of HNO3 in titration = Moles of KOH in the original solution
(0.500 mol/L HNO3) * (0.0223 L HNO3) = moles of KOH
0.01115 mol = moles of KOH
Now we can find the molar concentration of KOH using its moles and the volume of the original solution (21.15 mL):
Molarity of KOH = moles of KOH / volume of KOH solution in L
Molarity of KOH = 0.01115 mol / 0.02115 L = 0.527 mol/L
So the molar concentration of KOH in the original solution is 0.527 mol/L.
To find the molar concentration of KOH in the original solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between HNO3 and KOH:
HNO3 + KOH → KNO3 + H2O
From the balanced equation, we can see that the ratio between HNO3 and KOH is 1:1. This means that 1 mole of HNO3 reacts with 1 mole of KOH.
Given that the volume of the HNO3 solution used is 22.30 mL and the molar concentration of the HNO3 solution is 0.500 M, we can calculate the number of moles of HNO3 used:
moles of HNO3 = molar concentration × volume
= 0.500 M × 22.30 mL
= 11.15 mmol
Since the moles of HNO3 and KOH are equal, the number of moles of KOH is also 11.15 mmol.
Now, let's calculate the molar concentration of KOH in the original solution. The volume of the KOH solution used is 21.15 mL.
molar concentration of KOH = moles of KOH / volume of KOH solution
= 11.15 mmol / 21.15 mL
= 0.527 M
Therefore, the molar concentration of KOH in the original solution is 0.527 M.
To find the molar concentration of KOH in the original solution, we can use the concept of stoichiometry and the volume of titrant used in the titration.
First, we need to determine the number of moles of HNO3 used in the titration. We can do this by using the molarity (M) of the HNO3 solution and the volume (V) of the HNO3 solution used to reach the endpoint.
Molarity (M) is defined as moles of solute per liter of solution, so we can calculate the number of moles of HNO3 using the following equation:
moles of HNO3 = Molarity (M) x Volume (V in liters)
Given that the concentration (M) of the HNO3 solution is 0.500 M and the volume (V) used is 21.15 mL (which is equivalent to 0.02115 L):
moles of HNO3 = (0.500 M) x (0.02115 L) = 0.010575 moles
According to the balanced chemical equation between HNO3 and KOH, the stoichiometric ratio is 1:1. This means that for every 1 mole of HNO3, there is 1 mole of KOH.
Since the stoichiometric ratio is 1:1, the number of moles of KOH in the original solution is also 0.010575 moles.
Finally, we can calculate the molarity (M) of KOH in the original solution using the volume (V) used to reach the endpoint (22.30 mL or 0.02230 L):
Molarity (M) of KOH = moles of KOH / Volume (V in liters)
Molarity (M) of KOH = 0.010575 moles / 0.02230 L = 0.473 M
Therefore, the molar concentration of KOH in the original solution is 0.473 M.