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Step2247 [10]
2 years ago
15

A 10.0 mL sample of 0.75 M CH3CH2COOH(aq) is titrated with 0.30 M NaOH(aq) (adding NaOH to CH3CH2COOH). Determine which region o

n the titration curve the mixture produced is in, and the pH of the mixture at each volume of added base.Ka of CH3CH2COOH is 1.3 \times× 10−5.
Chemistry
1 answer:
Llana [10]2 years ago
3 0

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

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