CHEMISTRY HELP PLEASE!!! A 340.0 −mL buffer solution is 0.170 M in HF and 0.170 M in NaF.?

Refer to: http://clas.sa.ucsb.edu/staff/Resource%20folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf
pKa for HF = 3.14


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A.
Let y mol be the number of moles of NaOH added.

HF + NaOH → Na⁺ + F⁻ + H₂O
After reaction:
No. of moles of HF = (0.170 mol/L) × (340/1000 L) - (y mol) = (0.0578 - y) mol
No. of moles of F⁻ = (0.170 mol/L) × (340/1000 L) + (y mol) = (0.0578 + y) mol

[F⁻]/[HF] = (No. of moles of F⁻) / (No. of moles of HF) = (0.0578 + y) / (0.0578 - y)

The dissociation of HF :
HF(aq) ⇌ H⁺(aq) + F⁻(aq) …… pKa = 3.14

Henderson-Hasselbalch equation :
pH = pKa + log([F⁻]/[HF])
4.00 = 3.14 + log{(0.0578 + y) / (0.0578 - y)}
log{(0.0578 + y) / (0.0578 - y)} = 0.86
(0.0578 + y) / (0.0578 - y) = 10⁰˙⁸⁶
(0.0578 + y) / (0.0578 - y) = 7.244
0.0578 + y = 0.4187 - 7.244y
8.244y = 0.3609
y = 0.0438
0.0438 mol of NaOH is added.

Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol
Mass of NaOH added = (0.0438 mol) × (40.0 g/mol) = 1.75 g


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B.
Let Z mol be the number of moles of NaOH added.

HF + NaOH → Na⁺ + F⁻ + H₂O
After reaction:
No. of moles of HF = (0.350 mol/L) × (340/1000 L) - (z mol) = (0.119 - y) mol
No. of moles of F⁻ = (0.350 mol/L) × (340/1000 L) + (z mol) = (0.119 + y) mol

[F⁻]/[HF] = (No. of moles of F⁻) / (No. of moles of HF) = (0.119 + z) / (0.119 - z)

The dissociation of HF :
HF(aq) ⇌ H⁺(aq) + F⁻(aq) …… pKa = 3.14

Henderson-Hasselbalch equation :
pH = pKa + log([F⁻]/[HF])
4.00 = 3.14 + log{(0.119 + z) / (0.119 - z)}
log{(0.119 + z) / (0.119 - z)} = 0.86
(0.119 + z) / (0.119 - z) = 10⁰˙⁸⁶
(0.119 + z) / (0.119 - z) = 7.244
0.119 + z = 0.862 - 7.244y
8.244z = 0.743
z = 0.0901
0.0901 mol of NaOH is added.

Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol
Mass of NaOH added = (0.0901 mol) × (40.0 g/mol) = 3.60 g

My text shows the pKa of HF = 3.17. If your value for this is different, you will need to alter the calculations below.

Initial pH of the buffer = 3.17 because [HF] = [F-], so pH = pKa

Now, initial moles of each buffer component = 0.340 L X 0.170 mol/L = 0.0578 mol

To find the concentrations of each buffer component at pH = 4.0, know that as you add NaOH, you quantitatively convert HF into F-. So, letting x = moles NaOH added, and adding that to moles F- and subtracting that from moles F- gives:

pH = pKa + log [F-]/[HF]
4.0 = 3.17 + log (0.0578 + x) / (0.0578 -x)
6.76 = (0.0578+x) / (0.0578-x)
0.391 - 6.76x = 0.0578 +x
0.333 = 7.76x
x = 0.0429 mol NaOH
Mass NaOH = 0.0429 X 40.0 g/mol = 1.72 g NaOH