Calculate the molar concentration of a solution of nitric acid if 20.00 mL of the acid requires12.63 mL of 0.155 M NaOH to reach the equivalence point.
HNO3 + NaOH ==> NaNO3 + H2O
mols NaOH = M x L = ?
mols HNO3 = mols NaOH (Note 1 mol HNO3 = 1 mol NaOH in the balanced equation)
Then M HNO3 = mols HNO3/L HNO3
Well, it seems like you're in a pickle! But don't worry, I'm here to help you out... with a side of humor, of course!
To calculate the molar concentration of the nitric acid solution, we can use the equation:
M1V1 = M2V2,
where M1 is the concentration of the nitric acid, V1 is the volume of the nitric acid used, M2 is the concentration of the NaOH, and V2 is the volume of NaOH used.
Now let's substitute in our values:
M1 * 20.00 mL = 0.155 M * 12.63 mL.
Let's do some math magic to solve for M1:
M1 = (0.155 M * 12.63 mL) / 20.00 mL.
Calculating that gives us... drumroll please...
M1 = 0.0979875 M.
So, the molar concentration of the nitric acid solution is approximately 0.098 M. Ta-da!
Remember, if you ever need more assistance, just give me a holler!
To calculate the molar concentration of the nitric acid solution, we can use the concept of stoichiometry and the volume and concentration data provided.
1. Write the balanced chemical equation for the reaction between nitric acid (HNO3) and sodium hydroxide (NaOH):
HNO3 + NaOH ⟶ NaNO3 + H2O
2. Determine the moles of NaOH used:
Moles of NaOH = Volume of NaOH (in L) × Concentration of NaOH (in mol/L)
= 0.01263 L × 0.155 mol/L
= 0.00195565 mol
3. Since the balanced equation has a 1:1 stoichiometry between HNO3 and NaOH, the moles of HNO3 used will also be 0.00195565 mol.
4. Calculate the volume of HNO3 used by converting the given mL to L:
Volume of HNO3 = 20.00 mL = 0.02000 L
5. Calculate the molar concentration of HNO3:
Molar concentration of HNO3 = Moles of HNO3 / Volume of HNO3 (in L)
= 0.00195565 mol / 0.02000 L
= 0.09779 mol/L
Therefore, the molar concentration of the nitric acid solution is approximately 0.09779 mol/L.
To calculate the molar concentration of a solution of nitric acid, we can use the concept of stoichiometry and the equation of neutralization between nitric acid (HNO3) and sodium hydroxide (NaOH).
1. First, let's write the balanced chemical equation for the neutralization reaction between HNO3 and NaOH:
HNO3 + NaOH → NaNO3 + H2O
2. From the balanced equation, we can determine the stoichiometric ratio between the acid and the base. In this case, the mole ratio between HNO3 and NaOH is 1:1.
3. Now, let's convert the volume of NaOH used to moles. We know that the concentration of NaOH is 0.155 M, and the volume used is 12.63 mL.
Moles of NaOH = Concentration × Volume
Moles of NaOH = 0.155 M × 0.01263 L = 0.001949 Mols
4. Since the stoichiometric ratio of HNO3 to NaOH is 1:1, the moles of HNO3 used in the neutralization reaction are also 0.001949 moles.
5. We are given that 20.00 mL of nitric acid reacted with the NaOH. Since the stoichiometric ratio is 1:1, the volume of nitric acid is also 20.00 mL.
6. Finally, to calculate the molar concentration of nitric acid (HNO3), we use the formula:
Molarity = Moles of solute / Volume of solvent (in liters)
Molarity of HNO3 = 0.001949 moles / 0.02000 L
Molarity of HNO3 = 0.0975 M
Therefore, the molar concentration of the nitric acid solution is 0.0975 M.