in HNO3 titration, you add few drops of phenolphthalein indicator to 50.0 mL of acid in a flask. you quickly add 20.00 mL of 0.0547 M NaOH but overshoot the end point, and the solutions turn deep pink. instead of starting over, you add 30.00 mL of acid, and the solution turns colourless. then, it takes 7.05 mL of NaOH to reach the end point. what is the concentration of the HNO3 solution?
mols base = mols acid
M x L base = M x L acid
M x (add all the L base) = M x L (add on the L acid)
The only unknown is M acid.
To determine the concentration of the HNO3 solution, we need to calculate the number of moles of HNO3 and NaOH involved in the reaction. From there, we can use stoichiometry to find the ratio between the two substances and determine the concentration of HNO3.
Let's break down the information provided:
1. Initial volume of acid (HNO3) = 50.0 mL
2. Volume of NaOH added after overshooting end point = 20.00 mL
3. Volume of additional acid (HNO3) added = 30.00 mL
4. Volume of NaOH required to reach the end point = 7.05 mL
5. Concentration of NaOH solution = 0.0547 M
First, let's calculate the moles of NaOH used to reach the end point:
Moles of NaOH = (Volume of NaOH * Concentration of NaOH)
= (7.05 mL * 0.0547 M)
= 0.384 mmoles
Since the reaction between NaOH and HNO3 occurs in a 1:1 ratio, we know that the moles of NaOH used are equal to the moles of HNO3 in the solution.
Now, let's calculate the total moles of HNO3 present in the solution:
Moles of HNO3 = (Initial volume of acid + Additional acid) * Concentration of HNO3
Initial moles of HNO3 = (50.0 mL * Concentration of HNO3)
Additional moles of HNO3 = (30.00 mL * Concentration of HNO3)
Total moles of HNO3 = Initial moles of HNO3 + Additional moles of HNO3
Since the solution turned colorless after the addition of the additional acid, it means that the moles of HNO3 in the original solution are equal to the moles of HNO3 required to react with the initial volume of NaOH added (before overshooting the endpoint).
Therefore, Total moles of HNO3 = Moles of NaOH used to reach the endpoint
= 0.384 mmoles
Now, rearranging the equation for total moles of HNO3:
Total moles of HNO3 = Moles of HNO3 in the original solution
+ Moles of HNO3 in additional acid
0.384 mmoles = (50.0 mL * Concentration of HNO3)
+ (30.00 mL * Concentration of HNO3)
Solving for Concentration of HNO3:
Concentration of HNO3 = (0.384 mmoles) / (50.0 mL + 30.00 mL)
= (0.384 mmoles) / (80.0 mL)
= 0.0048 M
Therefore, the concentration of the HNO3 solution is 0.0048 M.