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What is the pH of a 0.35 M NaC7H5O2 solution? (KaHC7H5O2= 6.5x10-5)

User Slim Fadi
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1 Answer

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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.

User Kert
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