Question #48. In a titration experiment, 50 mL of 0.1 M acetic acid (Ka = 1.75...

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Question #48. In a titration experiment, 50 mL of 0.1 M aceticacid (Ka = 1.75 x 10-5) was titrated with a 0.1 M NaOH (formalconcentration) at 25 ÂºC. The pH of the solution at the equivalencepoint will be this: Please show work (1) 7.00; (2) 8.73; (3) 9.06;(4) None of the above

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4.3 Ratings (809 Votes)

answer : (2) 8.73

millimoles of acetic acid = 50 x 0.1 = 5

millimoles of base = 0.1 x V = 0.1V

at equivalence point millimoles of acid = millimoles of base

5 = 0.1 V

V = 50 mL

CH3COOH + NaOH ------------------> CH3COONa + H2O

Â Â Â Â Â Â Â 5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0

Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 0Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 5

here salt only remains . so

[salt] = millimoles of salt / total volume

Â Â Â Â Â Â Â Â = 5 / (50 + 50)

Â Â Â Â Â Â Â Â = 0.05 M

pH = 7 + 1/2 (pKa + log C)

Â Â Â Â Â = 7 + 1/2 (4.76 + log 0.05)

Â Â Â Â Â = 8.73

pH = 8.73

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