Respuesta :
Answer:
sln A: pH = 3.745
sln B: pH = 5.745
sln C: pH = 4.745
Explanation:
- CH3COOH ↔ CH3COO- + H3O+
- Ka = ( [ H3O+ ] * [ CH3COO- ] ) / [ CH3COOH ] = 1.8 E-5
sln A:
⇒ [ CH3COOH ] = 10 [ CH3COO- ]
⇒ [ CH3COO- ] / [CH3COOH ] = 1 / 10
⇒ Ka = 1.8 E-5 = [ H3O+ ] * (1 / 10)
⇒ [ H3O+ ] = (10) * ( 1.8 E-5 ) = 1.8 E-4
⇒ pH = - log ( 1.8 E-4 ) = 3.745
sln B:
⇒ [ CH3COO- ] = 10 [ CH3COOH ]
⇒ [ CH3COO- ] / [ CH3COOH ] = 10
⇒ Ka = 1.8 E-5 = 10 * [ H3O+ ]
⇒ [ H3O+ ] = 1.8 E-5 / 10 = 1.8 E-6
⇒ pH = - log ( 1.8 E-6 ) = 5.745
sln C:
⇒ [ CH3COO- ] = [ CH3COOH ]
⇒ [ CH3COO- ] / [ CH3COOH ] = 1
⇒ Ka = 1.8 E-5 = ( 1 ) * [ H3O+ ]
⇒ [ H3O+ ] = 1.8 E-5
⇒ pH = - log ( 1.8 E-5) = 4.745
We have that for the Question "Three acetic acid/acetate buffer solutions, A, B, and C, were made using varying concentrations: [acetic acid] ten times greater than [acetate], [acetate] ten times greater than [acetic acid], and [acetate]"
It can be said that
- [acetic acid] ten times greater than [acetate] = [tex]3.74[/tex]
- [acetate] ten times greater than [acetic acid] = [tex]5.74[/tex]
- [acetate]= [acetic acid] = [tex]4.74[/tex]
From the question we are told
Acetic acid has a Ka of [tex]1.8*10^{-5[/tex]
[tex]CH_3COOH ---------> CH_3COO^- + H^+[/tex]
for acetic acid and acetate buffer using Hendersom-Harsel Batch equation
[tex]p^H = p^ka + log \frac{CH_3COO^-}{CH_3COOH}\\\\p^{ka} = -log(ka)\\\\p^H = -log(ka) + log \frac{CH_3COO^-}{CH_3COOH}[/tex]
For acetic acid 10 times greater than acetate
[tex]\frac{CH_3COO^-}{CH_3COOH} = \frac{1}{10}\\\\p^H = -log (1.8*10^{-5}) + log\frac{1}{10}\\\\p^H = 3.74[/tex]
For acetate 10 times greater than acetic acid
[tex]\frac{CH_3COO^-}{CH_3COOH} = 10\\\\p^H = -log(1.8*10^{-5}) + log10\\\\p^H = 5.74[/tex]
For acetate=acetic acid
[tex]\frac{CH_3COO^-}{CH_3COOH} = 1\\\\p^H = -log(1.8*10^{-5}) + log1\\\\p^H = 4.74[/tex]
For more information on this visit
https://brainly.com/question/17756498
