Question

The Ka my textbook gives for benzoic acid is 6.28x10^-5.

Calculate the pH of the solution that results when 15.0 mL of 0.1800 M benzoic acid is (This problem requires values in your textbook's specific appendices, which you can access through the OWLv2 MindTap Reader. You should not use the OWLv2 References' Tables to answer this question as the values will not match.) a diluted to 60.0 mL with distilled water. pH = b mixed with 45.0 mL of 0.0600 M NaOH solution. pH = C mixed with 45.0 mL of 0.120 M NaOH solution. pH = d mixed with 45.0 mL of 0.120 M sodium benzoate solution. pH =

Answer 1

Benzoic acid 15.one 0.18M. diluted to 60 me with distilled water MVI : Mz vz before After dilution dilution IS Xo. 18 = M2 X 60 M2: 15 X 0.18 =6045 60 C: 0.005 ka = 6.28 x 105 THA ] = ſoka = ſo oust 6. 28X10 = 10.0282 X102 = 0.168 xronz pH = -logint) = -log(0.198 X102) 3 2 - 1 0.168 = 2+0.77 = 2.77 pH = 2.77 - b) mixed with us me of 0.06M Naoh ph coon' t / - Ph con Nat tho 0.18 M ismi Naon 0.06M 45 ml 2.7 mmol 2.7 mmol 2. 7 mmol Auo Baie Both acid and bare amount so are in solution same fi neutrilise and pH of is

+ phcoo nat tno ph coon 0.18M 15 ml Naon 0.12 M usmi 2.7 mm 5.4 mmol - - - (2.7 mmol 2.1 mmol Now volume will be i ust 15 = Gone Conch = 2.2 Go = 0.045 -0. ou SM Naon M Naon Tony = 0.ous POH = -logion'] = -log o.ous = 1.346 pH = 14-1.34 = 12.65 pr= 12.65 Mixed with us and of 0.12M sodium benzoate + sodium benzoate Butter Benzoic Acid - Weak acid Conjugate bat 0-12 x 45 = 2.7 mmol 5.4 monal رمان به (pH = pka + log salt pH = -log ka + log Site = -log (6.26 X155) + log 2 = 5 - reg 6.25 + log2 = 5-0.79 + 0.3 = 4.51 I PH=4.51