millimolwa HF = mL x M = 100 mL x 0.40 M = 40
millimoles LiOH = 100 x 0.2 M = 20
..................HF + LiOH ==> LiF + H2O
I................40..........0.............0.........0
add........................20............................
C...............-20.......-20............+20..................
E...............20..........0...............20.......................
So you have formed a buffer solution of HF and F (a weak acid and its salt in solution is a buffere). Use the Henderson-Hasselbalch equation to solve for pH. That's pH = pKa + log [(base)/(acid)]
You will need pKa. pKa = -log Ka for HF. You can find Ka in your text or on the webl.
You will need concn base (F^-) = 20 mmols/200 mL = ?
You will need concn acid (HF) = 20 mols/200 mL = ?
Post your work if you get stuck.
A 100 mL sample of 0.40 M hydrofluoric acid is mixed with 100.0 mL of 0.2M lithium hydroxide. Find the pH of the final solution. pKa for HF is 3.2.
3 answers
Would the equation for pH be pH=3.2+log[0.01/10] ?
(using the given pKa and molarity of the base and acid)
(using the given pKa and molarity of the base and acid)
yes and no.
pH = pKa + log (F^-)/(HF)
pKa is 3.2---yes
But your concentrations are not right.
(F^-) = 20 millimols/200 mL = 0.1
(HF) = 20 millimoles/200 mL = 0.1
pH = 3.2 + log (0.1/0.1) = 3.2 + log 1 = 3.2 + 0 = 3.2
pH = pKa + log (F^-)/(HF)
pKa is 3.2---yes
But your concentrations are not right.
(F^-) = 20 millimols/200 mL = 0.1
(HF) = 20 millimoles/200 mL = 0.1
pH = 3.2 + log (0.1/0.1) = 3.2 + log 1 = 3.2 + 0 = 3.2