Question

2. Calculate the pH at the following point in the titration of 20.00 mL of0.500 M CH3COOH with 0.500 M NaOH. CH3COOH has a Ka = 1.8x10-5. (2.5 pts) a. pH before the addition of any NaOH. Include balanced chemical equation. b. pH after the addition of 8.00 mL of 0.500 M NaOH. Include balanced chemical equation. c. pH after the addition of 10.00 mL of 0.500 M NaOH. Include balanced chemical equation. d. pH after the addition of 20.00 mL of 0.500 M NaOH. Include balanced chemical equation. e. pH after the addition of 21.00 mL of 0..500 M NaOH. Include balanced chemical equation.

Answer 1

a) pH = 2.52

b) pH = 4.56

c) pH = 4.74

d) pH = 9.069

e) pH = 12.086

(al of NaOH Chacao + H+ pH before addition CH₃Coon initial 0.5 final 0.5(1-0) 0.54 0.50 ka = (0.5%) (0.50 = 1.8x105 0.5(1-0) 0.542 = 118x105 - 1-ani x2 = 3,6x10s Q = 36x10-6 x = 6x10 3 = 0.006 x < 5 % So, the neglection is allowed CH+] = 0.50 = 0.580.006 -0.003 pH = -log Cht] = -log(0.003) [pH = 2.523]

(b) pH after addition of 8.00 mL o.5 M NaOH CH₃COONa + H2O t=0 milli moles CH₃COOH + NaOH 2oxois 8x0.5 104 t=t 6 - solution (CH₃COOH + CH₃ Coona) this booms a buffer H₂O is neutral. SO pH = pka + log [Criz cogna] Ссна собно pH = -log(1.8x105) + log 4/28 6/28 pH = 4.74 + (-0.17 609) (pH = 4.564)

(c) pH after the addition of 10 ml of Naoh CH₂COOH + NaOH - CH₃COONa + H₂O t=0 milli moles 20x0.5 loxois 10 t=t 5 Again, this forms a buffer solution pH = pka + log [CH₃COONa] [Cozcoor] pH = 4.74 + log($130 ) 5/30 pH = 4.74 + logt | PH = 4.74

(d) pH after the addition of 20 ml of ois M NOOH CH₃COOH + NaOH CH₃COONa + H₂O to 2oxois 20x0.5 milli moles 10 10 t-t - 10 10 weak acid and CH₃COONa is a salt formed from Stoong base SO, pH = 1 ( pkw + pka + logc) pka = 4.74 pkw = -log kw = -logloty [pkw = 14 c = concentration of salt c = 10 = 0.25 ५० So, pH = + (14+4.74 + logo125). 1 pH=9.069

ce) pH after the addition of 21.0 ml of 0.5 M Noon CH₃COOH + NaOH crezcooNa + H₂O to milli moles 20x0.5 218015 10.5 1o t=t 0.5 10 10 In this case ph d predominantly depends on NaoH Naos 0.5 Nat + oH to 0.5 0.5 concentration of COH) = moles total volume = 0.5 (20+21) = 0.012195 pOH = -log Con] = -log(0.012195) POH = 1.9138 pH = 14-poh = 14-1.9138 [PH= 12.0862]