pH = -log(HNO3)
-1.39 = log(HNO3)
I found HNO3 approximately 0.05 but you need to do it more accurately.
Then mol HNO3 = M x L = ?
mols HCl = M x L
total mols H^+ = mols HCl + mols HNO3.
volume = 145 mL + 493 mL = ?
M new solution = total mols/total liters.
Then pH = -log(H^+)
What is the pH of a solution obtained by adding 145mL of 0.575M HCl to 493mL of a HNO3 solution with a pH of 1.39?
1 answer
1 answer