Step-by-step explanation:
NaC₇H₅O₂ is the formula for sodium benzoate. It's a salt that is formed between a weak acid (benzoic acid) and a strong base (sodium hydroxide). When it is dissolved in water it will dissociate into its ions.
NaC₇H₅O₂ -----> Na⁺ + C₇H₅O₂⁻
The benzoate ion is the conjugate base of the benzoic acid. When dissolved in water it will act as a weak base.
C₇H₅O₂⁻ + H₂O <----> HC₇H₅O₂ + OH⁻ Kb = ?
We know the acid dissociation constant of the HC₇H₅O₂ (benzoic acid), but we need the base dissociation constant of C₇H₅O₂⁻ (benzoate). They are related by the Kw.
Ka = 6.5 * 10^(-5)
Ka * Kb = Kw
Kb = 1 * 10^(-14)/(6.5 *10^(-5))
Kb = 1.54 * 10^(-10)
C₇H₅O₂⁻ + H₂O <----> HC₇H₅O₂ + OH⁻ Kb = 1.54 * 10^(-10)
The expression for the Kb is:
Kb = [HC₇H₅O₂]*[OH⁻]/[C₇H₅O₂⁻] = 1.54 * 10^(-10)
The base dissociation constant gives us the relationship between the equilibrium concentrations of the different species. To find the equilibrium cconcentrations we have to set up an ICE table.
C₇H₅O₂⁻ + H₂O <----> HC₇H₅O₂ + OH⁻
I 0.35 0 0
C -x x x
E 0.35 - x x x
We can replace these values in the Kb expression and solve it for x.
[HC₇H₅O₂]*[OH⁻]/[C₇H₅O₂⁻] = 1.54 * 10^(-10)
x * x/(0.35 - x) = 1.54 * 10^(-10)
x² = 1.54 * 10^(-10) * (0.35 - x)
x² = 5.385 * 10^(-11) - 1.54 * 10^(-10) x
x² + 1.54 * 10^(-10) x - 5.385 * 10^(-11) = 0
We got a quadratic equation and we have to find its roots. They are:
x₁ = -0.000007338 x₂ = 0.000007338
We have to select the positive root because negative concentrations of compounds don't exist.
According to the ICE table the equilibrium concentration of OH- is equal to x. We can find the pOH with that value and finally the pH of our solution.
[OH⁻] = x
[OH⁻] = 0.000007338 M
pOH = - log [OH⁻]
pOH = - log 0.000007338 M
pOH = 5.13
pH = 14 - pOH
pH = 14 - 5.13
pH = 8.87
Answer: The pH of the solution is 8.87.