Answer:
[tex]pH=8.43[/tex]
Explanation:
Hello there!
In this case, since the reaction between benzoic acid and sodium hydroxide is carried out in a 1:1 mole ratio, it is possible to calculate the moles of sodium benzoate salt that are produced due to the neutralization as show below:
[tex]n_{salt}=0.2mol/L*1L=0.2mol[/tex]
We can assume that the volume of benzoic acid is 1 L and therefore the used volume of NaOH is:
[tex]V_{NaOH}=\frac{0.200M*1L}{0.060M}=3.33L[/tex]
Which means that the total final volume is 4.33 L and the concentration of the resulting basic salt is:
[tex][salt]=\frac{0.2mol}{4.33L}=0.046M[/tex]
Next, since the salt is C6H5COONa, we can see it is ionized in Na+ cations and C6H5COO- anions, but just the latter is able to react with the water in the solution in order to undergo the following reaction:
[tex]C_6H_5COO^-+H_2O\rightleftharpoons C_6H_5COOH+OH^-[/tex]
Whose equilibrium expression, it terms of Kb, due to the release of OH- ions, is:
[tex]Kb=\frac{[C6H5COOH][OH^-]}{[C6H5COO^-]}[/tex]
And can be written in terms of x, Kw and Ka:
[tex]\frac{Kw}{Ka} =\frac{x*x}{0.046M-x} \\\\\frac{1x10^{-14}}{6.3x10^{-5}}=1.59x10^{-10}= \frac{x^2}{0.046M-x}[/tex]
And because Kb<<<<<1 we can neglect the x on the bottom to get:
[tex]1.59x10^{-10}= \frac{x^2}{0.046M}[/tex]
Whereas x is computed as follows:
[tex]x=\sqrt{1.59x10^{-10}*{0.046M} =2.7x10^{-6}M[/tex]
Which is actually equal to the concentration of OH- ions so that we can calculate the pOH prior to the pH:
[tex]pOH=-log(2.7x10^{-6})=5.57[/tex]
And therefore the pH as shown below:
[tex]pH=14-5.57\\\\pH=8.43[/tex]
Which makes sense since this sodium benzoate is a basic salt.
Regards!
