Question

A chemist titrates of a butanoic acid solution with solution at . Calculate the pH at equivalence. The of butanoic acid is__________ .Round your answer to decimal places.

Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of solution added.

Answer 1

The pH at equivalence for the titration of a butanoic acid solution with sodium hydroxide
`(\(NaOH\))` at
`\(0.1 \ M\)` is approximately 8.21. The
`\(K_a\)` value for butanoic acid is
`\(1.54 * 10^(-5)\)`.

Step-by-step explanation:

In the titration of butanoic acid
`(\(CH_3CH_2CH_2COOH\))` with sodium hydroxide
`(\(NaOH\))`, at the equivalence point, the moles of
`\(OH^-\)` ions added equal the moles of
`\(H^+\)` ions present in butanoic acid. The chemical equation for this reaction is:

`\[CH_3CH_2CH_2COOH + NaOH \rightarrow CH_3CH_2CH_2COONa + H_2O\]`

At the equivalence point, the butanoic acid is completely neutralized, and the resulting solution is the sodium salt of butanoic acid. The pH at equivalence can be calculated using the
`\(K_a\)` expression for butanoic acid:

`\[K_a = ([CH_3CH_2CH_2COO^-][H_3O^+])/([CH_3CH_2CH_2COOH])\]`

Since we're at the equivalence point, the concentration of
`\(CH_3CH_2CH_2COOH\)` is zero, and we can simplify the expression to:

`\[K_a = ([H_3O^+])/([CH_3CH_2CH_2COO^-])\]`

Rearranging the equation to solve for
`\([H_3O^+]\)`, we get:

`\[ [H_3O^+] = K_a * [CH_3CH_2CH_2COO^-]\]`

Substituting the
`\(K_a\)` value for butanoic acid
`(\(1.54 * 10^(-5)\))`, and considering that at the equivalence point the concentration of
`\(CH_3CH_2CH_2COO^-\)` is the same as the concentration of
`\(NaOH\)` added, we can calculate the pH:

`\[pH = -\log[H_3O^+]\]`

`\[pH = -\log(K_a * [NaOH])\]`

`\[pH = -\log(1.54 * 10^(-5) * 0.1)\]`

`\[pH \approx 8.21\]`

Therefore, the pH at equivalence is approximately
`\(8.21\)`, and this result indicates a slightly basic solution.

Answer 2

pH = 8.75

Step-by-step explanation:

100.0mL of a 0.8108M of a butanoic acid (HC₃H₇CO₂, pKa 4.82) solution is titrated with 0.0520M KOH.

The reaction is:

HC₃H₇CO₂ + KOH → H₂O + KC₃H₇CO

Moles of butanoic acid are:

0.1000L × (0.8108mol / L) = 0.08108 moles of butanoic

For a complete reaction, volume of KOH must be added is the volume in which 0.08108 moles of KOH are added, that is:

0.08108 mol × (L / 0.0520mol) = 1.56L of KOH.

Total volume in equilibrium is 1.56L + 0.10L = 1.66L

That means concentration of butanoic acid is:

0.08108 mol / 1.66L = 0.04884M HC₃H₇CO₂

At equivalence point, there is just C₃H₇CO⁻ in solution

Kb of butanoic acid is:

C₃H₇CO⁻ + H₂O ⇄ HC₃H₇CO₂ + OH⁻

Kb = Kw / Ka

Ka = 10^-pKa

Ka = 1.51x10⁻⁵

Kb = 1x10⁻¹⁴ / 1.51x10⁻⁵ = 6.61x10⁻¹⁰

The equilibrium of Kb is:

Kb = 6.61x10⁻¹⁰ = [HC₃H₇CO₂] [OH⁻] / [C₃H₇CO⁻]

As at equivalence point there is just C₃H₇CO⁻, the equilibrium concentrations are:

[C₃H₇CO⁻] = 0.04884M - X

[HC₃H₇CO₂] = X

[OH⁻] = X

Replacing in Kb:

6.61x10⁻¹⁰ = X² / [0.04884M - X]

0 = X² + 6.61x10⁻¹⁰X - 3.23x10⁻¹¹

Solving for X:

X = -5.68x10⁻⁶ → False solution. There is no negative concentrations

X = 5.683x10⁻⁶ → Right solution.

As [OH⁻] = X, [OH⁻] = 5.683x10⁻⁶.

pOH = - log [OH⁻]

pOH = 5.245

pH = 14 - pOH

pH = 8.75