Calculate the pH and the pOH for each of the following solutions:

Question

1. A solution made by adding 4.67 grams of potassium hydroxide to enough water to make 5.00 L of final solution.
2.A solution made by adding 50 mL of 6.0 M hydrochloric acid to enough water to produce 2.0 L of final solution.
3. A solution made by adding 30.0 mL of 1.00 M nitric acid to 100 mL of water.Assume additive volum

Answer 1

1.mols KOH = grams/molar mass = 4.67/56.1 = 0.0832
M KOH = mols/L solution = 0.0832/5.00 = ? = (KOH) = (OH^-)
pOH = -log (KOH) = ?
Then pOH + pH = pKw = 1E-14. You know Kw and pOH, solve for pH.

2. 6 M HCl is being diluted so use the dilution formula.
mLa x Ma = mLb x Mb
50 mL x 6 M = 2000 mL x Mb
Solve for Mb which gives you the concentration of the "new" HCl. Then convert to pH from there since (H^+) = (HCl). Convert that to pOH.
3. Done the same way as #2.
NOTE: Since all of these (KOH, HCl, HNO3) are strong electrolytes, the (OH^-) or (H^+) is the same as (KOH) or (HCl) or (HNO3).
Post your work if you get stuck.