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A 360.0 −mL buffer solution is 0.130 M in HF and 0.130 M in NaF. Part...

A 360.0 −mL buffer solution is 0.130 M in HF and 0.130 M in NaF.

Part A) What mass of NaOH can this buffer neutralize before the pH rises above 4.00?

Part B) If the same volume of the buffer was 0.360

M in HF and 0.360 M in NaF, what mass of NaOH could be handled before the pH rises above 4.00?

Expert Solution

part A

1.3896g

Explanation

i)At equal concentration of HF and F^- , pH will be 3.17(pKa of HF)

Initial moles of HF = (0.130moles/1000ml)×360ml = 0.0468mol

Initial moles of F^- = (0.130moles/1000ml)×360ml = 0.0468mol

ii) Henderson-Hasselbalcch equation is as follows

pH = pKa + log([A^-]/[HA])

if pH is 4

4 = 3.17 + log ([A^-]/[HA])

log([A^-] /[HA]) = 0.83

[A^-]/[HA]= 6.761

[F^-] /[HF] = 6.761

moles of F^-/moles of HF = 6.761

moles of F^-= 6.761 moles of HF

iii) Total moles of buffer = 0.0468mol + 0.0468mol = 0.0936mol

at pH = 4

moles of HF + 6.761 moles of F^- = 0.0936mol

7.761moles of HF = 0.0936mol

moles of HF= 0.01206mol

moles of F^- = 0.0936mol - 0.01206mol = 0.01854mol

iv) NaOH react with HF to bring the pH above 4

moles of HF at pH 3.17 = 0.0468mol

moles of HF at pH 4.00 = 0.01206mol

moles of HF consumed = 0.0468mol - 0.01206mol = 0.03474mol

Moles of NaOH required = 0.03474mol

molar mass of NaOH = 40g/mol

mass of NaOH to be added = 0.03474mol × 40g/mol = 1.3896g

part B

i) Initial moles of F^-= (0.360mol/1000ml)×360ml = 0.1296mol Initial moles of HF = (0.360mol/1000ml) ×360ml = 0.1296mol

ii)

3.848g

Explanation

at pH = 4

moles of F^-/moles of HF = 6.761

moles of F^- = 6.761 moles of HF

Total moles = 0.1296mol + 0.1296mol = 0.2592mol

moles of HF + 6.761moles of HF = 0.2592mol

7.761moles of HF = 0.2592mol

moles of HF = 0.0334mol

iii) NaOH react with HF

moles of HF at pH 3.17 = 0.1296mol

moles of HF at pH 4.00 = 0.0334mol

moles of HF consumed = 0.0962mol

moles of NaOH to be added = 0.0962mol

mass of NaOH to be added = 3.848

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