Respuesta :
Answer:
5.75
Explanation:
Weak acid and strong base,
Make it a two part problem
The first will be Partial neutralization of the weak acid, while the second is the equilibrium and final pH
1.Parameters
HC2H3O2 =10mL x 0.75M = 7.5 mmol
NaOH =22mL x 0.30M = 6.6 mmol
R HC2H3O2 + OH- --> C2H3O2- + H2O
I 7.5 mmol 6.6 mmol 0 mmol ignore
C -6.6 mmol -6.6 mmol +6.6 mmol
E 0.9 mmol 0.0 mmol 6.6 mmol
2.) HC2H3O2 --> H+ + C2H3O2-
Initial concentrations
[HC2H3O2]: 0.9 mmol / 32mL = 0.0281M
[H+]:
[C2H3O2-]: 6.6 mmol x 32mL = 0.206M
Concentrations at equilibrium
[HC2H3O2]: 0.0281M - x
[H+]: [C2H3O2-]: 0.206M + x
If x is small and can be ignored except [H+]
Substitute and solve
1.3 x 10-5 = (x)(0.206)
(0.0281)
X= 1.77 x 10-6M => pH = 5.75
All acid titration curves follow the same basic shapes. In the beginning, the solution has a low pH and climbs as the strong base is added. The pH of the mixture at each volume of added base is 5.75.
Neutralization of acid:
A neutralization reaction is when an acid and a base react to form water and a salt.
Given:
HC₂H₃O₂ =10mL * 0.75M = 7.5 mmol
NaOH =22mL * 0.30M = 6.6 mmol
  R HC₂H₃O₂ + OH- --> C₂H₃O₂- + H₂O
I   I7.5 mmol   6.6 mmol  0 mmol
C  -6.6 mmol  -6.6 mmol  +6.6 mmol
E   0.9 mmol   0.0 mmol 6.6 mmol
The dissociation of acid is represented as:
HC₂H₃O₂- --------> H+ + C₂H₃O₂-
Initial concentrations
[HC₂H₃O₂]: 0.9 mmol / 32mL = 0.0281M
[C₂H₃O₂-]: 6.6 mmol * 32mL = 0.206M
Concentrations at equilibrium:
[HC₂H₃O₂]: 0.0281M - x
[H+]: [C₂H₃O₂-]: 0.206M + x
If x is small and can be ignored except [H+]
Substitute and solve
1.3 * 10-5 = (x)(0.206)
[tex]x= 1.77 *10^{-6} M[/tex]
pH = 5.75
Find more information about pH here:
brainly.com/question/22390063
