Question

A flask contains 250.0 mL of nitric acid (HNO3) at an unknown
concentration. Nitric acid is a strong acid and dissociates completely
into its ions when mixed with water.
o 10.0 mL of the original solution is diluted up to 100 mL to give solution
#2.
o 15.0 mL of solution #2 is diluted up to 100.0 mL to form solution #3.
o 5.00 mL of solution #3 is diluted up to 100.0 mL to form solution #4.
o 25.0 mL of solution #4 has a pOH of 10.18.
a. What is the concentration of nitric acid, or [HNO3], in the original
solution?
b. What is the pH of the original solution of nitric acid?

Answer 1

Because pH + pOH = 14, pH of solution #4 is 3.82.

Then concentration on acid is 10^-3.82
Use v1c1 = v2c2 to calculate concentrations of stronger solutions.
E.g. Concentration on solution #3 is 20 times greater.

Finally, pH = - log c

Answer 2

Thank you for the input! I appreciate it, however, I'm still pretty confused as to follow through. Can you run it to me step by step as to how you get the final answer, thanks again!

Answer 3

I sent you a message about that.

Did you get [H3O+] = 10^-3.82 mol/l = 0.000151356 mol/l
For solution #4?

The volume mentioned 25.0 ml is not important in this case.