Question

A 1.30 L buffer solution consists of 0.107 M butanoic acid and 0.345 M sodium butanoate. Calculate the pH of the solution following the addition of 0.075 moles of NaOH . Assume that any contribution of the NaOH to the volume of the solution is negligible. The Ka of butanoic acid is 1.52×10−5 .

Answer 1

The pH of the solution following the addition of 0.075 moles of NaOH is 5.7.

Step-by-step explanation:

The equation of the buffer solution is the following:

C₄H₇O₂H(aq) + H₂O(l) ⇄ C₄H₇O₂⁻(aq) + H₃O⁺(aq) (1)

The pH of the buffer solution can be found using the Henderson-Hasselbalch equation:

`pH = pKa + log(([NaC_(4)H_(7)O_(2)])/([C_(4)H_(8)O_(2)]))` (2)

The NaOH added will react with butanoic acid tot produce sodium butanoate:

C₄H₇O₂H(aq) + OH⁻(aq) ⇄ C₄H₇O₂⁻(aq) + H₂O(l) (3)

`n_{C_(4)H_(7)O_(2)H}_(i) = C*V = 0.107 M*1.30 L = 0.139 moles`

`n_{C_(4)H_(7)O_(2)^(-)} = 0.345 M*1.30 L = 0.449 moles`

`n_(NaOH) = 0.075 moles`

`n_{C_(4)H_(7)O_(2)H} = 0.139 moles - 0.075 moles = 0.064 moles`

`n_{C_(4)H_(7)O_(2)^(-)} = 0.449 moles + 0.075 moles = 0.524 moles`

`C_{C_(4)H_(7)O_(2)H} = (0.064 moles)/(1.30 L) = 0.049 M`

`C_{C_(4)H_(7)O_(2)^(-)} = (0.524 moles)/(1.30 L) = 0.403 M`

Now, from equation (1) we have:

C₄H₇O₂H(aq) + H₂O(l) ⇄ C₄H₇O₂⁻(aq) + H₃O⁺(aq)

0.049 - x 0.403 + x x

`Ka = ([C_(4)H_(7)O_(2)^(-)][H_(3)O^(+)])/(C_(4)H_(7)O_(2)H)`

`1.52 \cdot 10^(-5) = ((0.403 + x)x)/(0.049 - x)`

`1.52 \cdot 10^(-5)*(0.049 - x) - (0.403 + x)x = 0`

By solving the above equation for x we have:

x = 1.85x10⁻⁶ = [H₃O⁺]

So, the concentration of butanoic acid and sodium butanoate is:

`[C_(4)H_(7)O_(2)^(-)] = 0.345 + 1.85 \cdot 10^(-6) = 0.345 M`

`[C_(4)H_(7)O_(2)H] = 0.049 - 1.85 \cdot 10^(-6) = 0.049 M`

Finally, from equation (2) we have:

`pH = pKa + log(([NaC_(4)H_(7)O_(2)])/([C_(4)H_(8)O_(2)]))`

`pH = -log(1.52 \cdot 10^(-5)) + log((0.345)/(0.049)) = 5.7`

Therefore, the pH of the solution following the addition of 0.075 moles of NaOH is 5.7.

I hope it helps you!