1.Let's call the weak acid HA, then (HA) in mols/L = 0.0017 moles/0.500L = 0.0034 M. Then
..................HA ==> H^+ + A^-
I................0.0034..........0.....0
C..................-x.............x......x
E............0.0034 - x.......x......x
pH = -log (H^+) = 5.78. Solve for (H^+). It will be approximately 1E-6 and that will be x but you should use the more exact answer.
Substitute the E line (equilibrium) in to the Ka expression and solve for Ka. You will need to use the quadratic equation. You may want to think about correcting for the H^+ from H2O but that won't amount to more than about 5% error and I wouldn't try to correct for that. I don't think it's worth the effort since the Ka values are not known any better than that.
Can someone solve these 2 problems for me?
1. When 0.0017 moles of a weak acid are dissolved in enough water to make 0.500 L, the pH of the resulting solution is 5.78. Calculate Ka.
2. A solution is made by dissolving 15.5 g of NaOH in approximately 450 mL of water in a volumetric flask. The solution becomes quite warm, but after it is allowed to return to room temperature, water is added to total 500 mL of solution. Calculate the pH of the final solution. Report pH to 2 decimal places.
2 answers
2. On both 1 and 2 please post your work if you get stuck.
For 2 you have 15.5 g NaOH in 500 mL solution which is 15.5 g/40 g/mol = 0.3875 mols/0.500 L = 0.775 M
So NaOH is strong base, it ionizes completely so (OH^-) will be 0.775 M.
Convert that to pOH then convert to pH using
pH + pOH = pKw = 14. You know pKw and pOH so solve for pH.
pOH = about 0.1 so pH = about 13.9
For 2 you have 15.5 g NaOH in 500 mL solution which is 15.5 g/40 g/mol = 0.3875 mols/0.500 L = 0.775 M
So NaOH is strong base, it ionizes completely so (OH^-) will be 0.775 M.
Convert that to pOH then convert to pH using
pH + pOH = pKw = 14. You know pKw and pOH so solve for pH.
pOH = about 0.1 so pH = about 13.9
1 answer