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.

pH = 8.75

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

chemists can use the electron cloud model to assign electrons to different atomic orbitals. atomic orbitals also explain the patterns in the periodic table

explanation:

answer: e.   sour taste

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bases are those substances which dissociates in water to give ions or accepts ions. as they dissociate to give ions, they act as electrolytes and thus are electrically conductive. bases are bitter in taste. bases are slippery in nature.