Suppose that 10.9 mL of HNO3 is neutralized by 83.9 mL of a 0.0021 < solution of KOH in a titration. Calculate the concentration of the HNO3 solution.
Answer in units of M
To calculate the concentration of the HNO3 solution, we need to use the concept of stoichiometry and the information provided in the titration.
First, let's determine the number of moles of KOH used in the reaction. We can do this by multiplying the volume (in liters) of the KOH solution by its concentration:
moles of KOH = volume of KOH solution (in L) × concentration of KOH solution (in M)
Given that the volume of the KOH solution is 83.9 mL and its concentration is 0.0021 M, we convert the volume to liters:
volume of KOH solution = 83.9 mL = 83.9 mL × (1 L / 1000 mL) = 0.0839 L
Now, we can calculate the moles of KOH used:
moles of KOH = 0.0839 L × 0.0021 M = 0.00017619 mol
Using the balanced chemical equation for the reaction between HNO3 and KOH, we can see that the stoichiometric ratio is 1:1. Therefore, the moles of HNO3 used in the reaction are also equal to 0.00017619 mol.
Next, we calculate the concentration of the HNO3 solution. We divide the moles of HNO3 by the volume of HNO3 solution:
HNO3 concentration = moles of HNO3 / volume of HNO3 solution (in L)
Given that the volume of the HNO3 solution is 10.9 mL, we convert the volume to liters:
volume of HNO3 solution = 10.9 mL = 10.9 mL × (1 L / 1000 mL) = 0.0109 L
Now, we can calculate the concentration of the HNO3 solution:
HNO3 concentration = 0.00017619 mol / 0.0109 L = 0.01615 M
Therefore, the concentration of the HNO3 solution is 0.01615 M.
To calculate the concentration of the HNO3 solution, we need to use the concept of molarity (M), which is defined as the number of moles of solute per liter of solution.
First, let's find the number of moles of KOH used in the neutralization reaction. We can use the equation:
n(KOH) = M(KOH) × V(KOH)
where n(KOH) is the number of moles of KOH, M(KOH) is the molarity of the KOH solution, and V(KOH) is the volume of the KOH solution in liters.
Given that the volume of the KOH solution is 83.9 mL (or 0.0839 L) and the molarity of the KOH solution is 0.0021 M, we can plug these values into the equation to find the number of moles of KOH:
n(KOH) = 0.0021 M × 0.0839 L
n(KOH) ≈ 0.00017559 moles
Next, let's determine the mole ratio between KOH and HNO3 in the neutralization reaction, which is 1:1 based on the balanced chemical equation:
HNO3 + KOH → KNO3 + H2O
Since the mole ratio is 1:1, the number of moles of HNO3 used in the neutralization reaction is also 0.00017559 moles.
Finally, we can calculate the concentration of the HNO3 solution using the equation:
M(HNO3) = n(HNO3) / V(HNO3)
where M(HNO3) is the concentration of the HNO3 solution, n(HNO3) is the number of moles of HNO3, and V(HNO3) is the volume of the HNO3 solution in liters.
Given that the volume of the HNO3 solution is 10.9 mL (or 0.0109 L), we can plug these values into the equation to find the concentration of the HNO3 solution:
M(HNO3) = 0.00017559 moles / 0.0109 L
M(HNO3) ≈ 0.01613 M
Therefore, the concentration of the HNO3 solution is approximately 0.01613 M.