(1a) What is the pH of a bufferthat contains 0.150 M acetic acid and 0.250 M sod

Question

(1a)What is the pH of a bufferthat contains 0.150 Macetic acid and 0.250 Msodium acetate?
(1b)What is the pH of the bufferin part1a after 0.0500 moles of HClis added to 1.00 L of the buffer?
(1c) What is the pH of thebuffer in part 1aafter 50.0 mL of 0.100M KOH is added to 500. mL ofthe buffer?
(1d) What is the pH of500. mL of water after 50.0 mL of 0.100 MKOH is added to it? (1a)What is the pH of a bufferthat contains 0.150 Macetic acid and 0.250 Msodium acetate?
(1b)What is the pH of the bufferin part1a after 0.0500 moles of HClis added to 1.00 L of the buffer?
(1c) What is the pH of thebuffer in part 1aafter 50.0 mL of 0.100M KOH is added to 500. mL ofthe buffer?
(1d) What is the pH of500. mL of water after 50.0 mL of 0.100 MKOH is added to it?

Explanation / Answer

Part A:             Formula:                              pH = pKa + log [salt] / [acid]      For acetic acid pKa = 4.74                            [salt] = 0.250 M                           [acid] = 0.150 M                           pH = 4.74 + log ( 0.250 / 0.150)                                     = 4.96    Part B:                No.of moles of salt = 0.250 M * 1L                                               =0.250 moles                No.of moles of acid = 0.150 M * 1.00 L                                               = 0.150 moles                 No.of moles of acid = 0.05 moles When a strong acid like HCl is added to the acidic buffer ,no.of moles of acid increases and number of moles of saltdecreases.                  No.of moles of salt = 0.250 moles - 0.05 moles                                                = 0.2 moles                 No.of moles of acid =   0.150 moles + 0.05 moles                                                = 0.2 moles                        Asboth are equal pH = pKa                                    pH = 4.74 Part C:             When a strong base is added to the acidic buffer number of moles ofsalt increases and number of moles of acid decreases.                  No.ofmoles of salt = 0.250 M * 0.5L                                               =0.125 moles                No.of moles of acid = 0.150 M * .500 L                                               = 0.075 moles            No.of moles of NaOH = 0.005 moles                  No.ofmoles of salt = 0.125 moles +0.005 moles                                               = 0.13 moles                No.of moles of acid = 0.075 moles - 0.005 mols                                               = 0.07 moles                                    pH   = 4.74 + log ( 0.13 / 0.07 )                                               = 5.00 d) [OH-] = 0.05 L * 0.1 M / 0.55L                 = 0.009 M         pOH = 2.04          pH = 11.95 Part C:             When a strong base is added to the acidic buffer number of moles ofsalt increases and number of moles of acid decreases.                  No.ofmoles of salt = 0.250 M * 0.5L                                               =0.125 moles                No.of moles of acid = 0.150 M * .500 L                                               = 0.075 moles            No.of moles of NaOH = 0.005 moles                  No.ofmoles of salt = 0.125 moles +0.005 moles                                               = 0.13 moles                No.of moles of acid = 0.075 moles - 0.005 mols                                               = 0.07 moles                                    pH   = 4.74 + log ( 0.13 / 0.07 )                                               = 5.00 d) [OH-] = 0.05 L * 0.1 M / 0.55L                 = 0.009 M         pOH = 2.04          pH = 11.95                                               =0.125 moles                No.of moles of acid = 0.150 M * .500 L                                               = 0.075 moles            No.of moles of NaOH = 0.005 moles                  No.ofmoles of salt = 0.125 moles +0.005 moles                                               = 0.13 moles                No.of moles of acid = 0.075 moles - 0.005 mols                                               = 0.07 moles                                    pH   = 4.74 + log ( 0.13 / 0.07 )                                               = 5.00 d) [OH-] = 0.05 L * 0.1 M / 0.55L                 = 0.009 M         pOH = 2.04          pH = 11.95              

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