Question

A 1.50L buffer solution is 0.250 M in HF and 0.250 in NaF. Calculate the ph of the solution after the addition of 0.0500 moles of solid NaOH. Assume no volume change upon the addition of the base. The Ka for HF is 3.5E^-4

Answer 1

The pH is 3.57

Step-by-step explanation:

Step 1: Data given

Volume = 1.50 L

Molarity of HF = 0.250 M

Molarity of NaF = 0.250 M

addition of 0.05 moles NaOH

Step 2:

The buffer contains HF, which is a weak acid; and NaF, which is the salt of its conjugate base, the fluoride anion, F.

−

When adding NaOH ( and colume will not change), we expect the weak acid to neutralize the strong base (and vice versa).

HF(aq) + NaOH(aq) → NaF(aq) + H2O(l)

For 1 mole HF consumed, we need 1 mole NaOH, to produce 1 mole NaF and 1 mole H2O

Step 3: Calculate moles of HF

Moles HF = Molarity HF * Volume HF

Moles HF = 0.250 M * 1.5 L

Moles HF = 0.375 moles HF

Step 4: Calculate moles of F- (conjugate base)

Moles F- = 0.250 M * 1.5 L

Moles F- = 0.375 moles F-

Step 5: Addition of 0.05 moles NaOH

HF + OH- → F- + H2O

Since we have fewer moles of strong base than on weak acid, the sodium hydroxide will be completely consumed.

The number of moles HF will change to:

Moles HF = 0.375 - 0.05 = 0.325 moles HF

The number of moles F- will change to:

Moles F- = 0.375 + 0.05 = 0.425 moles F-

Step 6: Calculate molarity

Molarity HF = Moles HF / Volume

Molarity HF = 0.325/1.5 = 0.21667 M

Molarity F- = 0.425/1.5 = 0.28333 M

Step 7: Calculate pKa

3.5 * 10^-4 = Ka pKa = - log(3.5 * 10^-4 ) = 3.456

Step 8: Calculate pH

pH = - log(3.5 * 10^-4 ) + log[F-]/[HF]

pH = 3.456 + log (0.28333/0.21667)

pH = 3.57

The pH is 3.57