Question

please help

Calculate the pH of the solution at every stage of titration. The Ka of CH3COOH is 2.4 x10-5. 2 decimal places for all three answers in this question. a. After the addition of 20.0 ml of 0.10 M NaOH to 20.0 ml of 0.15 M CH3COOH. b. After the addition of 30.0 ml of 0.10 M NaOH to 20.0 ml of 0.15 M CH3COOH. c. After the addition of 40.0 ml of 0.10 M NaOH to 20.0 ml of 0.15 M CH3COOH.

Answer 1

NOTE : THIS ANSWER IS BIG AND LENGTHY.

Answers

a). 4.92

b). 8.70

c). 13.32

Details calculation given in the attachment....

Answer: a. 4.92 b. 8.7 C. 13.32 Calculation : given, Ka of CH₂COOH = 2.4x105 for every stage of 0.15 M CH₂ cool 20 ml 1 L=1000 mL) 0.02 L so 20 ml = Now, (20 ml = 0.02L) addition of 20. me of Calculate moles of CH₂ Cool 0.10 M NaOH and Nach number of moles = Volume x molarity Number of moles of CH₂ Cook 0.003 moles CH₃COOH Number of moles of NaOH = 220.0 0 2 moles Nach = -al m) x 6.024

+ H₂O (1) CH₂COOH 0.003 + OH 0.003 - e CH₂ cool o Before +0.003 Change -0.003 -0.003 after o 0.003 Both Now are fully nutralize the reaction so Concentration of CH₃ CooNa o.o6 M 0.003 moles O.OSL salt Now we know Hydrolysis for weak acid and strong Base pH = + [pkw tpka + log c] where pkw is water dissociation Constant = pka = 4.62 log c = log (0-06M) 14 so pH = + [14 + 4.62 + log 6.00m)] pH a b [14+ 4.62 + (-1.22)] pH = 8. 7 Answer