Question
How many grams of nitric acid (HNO3) must be dissolved in 590 ml of water to make a solution with a pOH = 10.1?
Select one:
a. 2.9 x 10-9 g
b. 7.9 x 10-3 g
c. 4.7 x 10-3 g
d. 7.9 x 10-6 g
e. 7.4 x 10-5 g
Select one:
a. 2.9 x 10-9 g
b. 7.9 x 10-3 g
c. 4.7 x 10-3 g
d. 7.9 x 10-6 g
e. 7.4 x 10-5 g
Answers
DrBob222
Convert pOH to pH with
pH + pOH = pKw = 14.
Convert pH to (H^+) with pH = -log(H^+).
Then since HNO3 is 100% ionized, H^+) = (HNO3)
mols HNO3 needed = M x L = ?
mols HNO3 = grams/molar mass. You know mols and L, solve for grams.
pH + pOH = pKw = 14.
Convert pH to (H^+) with pH = -log(H^+).
Then since HNO3 is 100% ionized, H^+) = (HNO3)
mols HNO3 needed = M x L = ?
mols HNO3 = grams/molar mass. You know mols and L, solve for grams.