( 2a ) What is the pH of1.00 L of a buffer that is 0.250 M in CH3NH2 and0.250 M

Question

(2a) What is the pH of1.00 L of a buffer that is 0.250 Min CH3NH2 and0.250 M CH3NH3Cl?
(2b) What is the pH of thebuffer in part 2aafter 25.0 mmole of HBr is added to 1.00 Lof the buffer? (2c) What is the pH of thebuffer in part 2aafter 100. mL of 0.0100M NaOH is added to 1.00 L ofthe buffer?
(2d) What is the pH ofwater after 100. mL of 0.0100 MNaOH is added to 1.00 L of water? Does thisillustrate buffer action? (2a) What is the pH of1.00 L of a buffer that is 0.250 Min CH3NH2 and0.250 M CH3NH3Cl?
(2b) What is the pH of thebuffer in part 2aafter 25.0 mmole of HBr is added to 1.00 Lof the buffer? (2c) What is the pH of thebuffer in part 2aafter 100. mL of 0.0100M NaOH is added to 1.00 L ofthe buffer?
(2d) What is the pH ofwater after 100. mL of 0.0100 MNaOH is added to 1.00 L of water? Does thisillustrate buffer action?

Explanation / Answer

Formula:                 pH = 14 - pOH             pOH = pKb + log [salt] / [base]              pKb  =  3.35            [salt]  = 0.250 M            [base] = 0.250 M As both are of same concentrations, pOH = pKb            pOH = 3.35 + log ( 0.250 / 0.250 )                      =3.35             pH   = 14 - 3.35                      = 10.65 Part B:                  When a strong acid like HBr is added to the basic buffer , numberof moles of salt increase and number of moles of basedecreases.    Number of moles of salt = 0.250 moles + 0.025 moles                                         =  0.275moles    Number of moles of base = 0.250 moles -0.025 moles                                           = 0.225 moles
                                pOH = 3.35 + log ( 0.275 / 0.225 )                                          = 3.43                                       pH = 10.57 Part C:                  Whena strong base like NaOH is added to the basic buffer ,number of moles of salt decrease and number of moles of baseincreases. No.of moles of NaOH = 0.001 moles Number of moles of salt = 0.250moles -  0.001 moles                                         =  0.249moles    Number of moles of base = 0.250 moles +0.001 moles                                           = 0.251 moles
                              pOH   = 3.35 + log ( 0.249 / 0.251 )                                         =  0.249moles    Number of moles of base = 0.250 moles +0.001 moles                                           = 0.251 moles
                                         = 3.34                                     pH= 10.66 Part D:                                [OH-] = 0.1 L * 0.01 M / 1.1 L                                           = 9.09 * 10-4 M                               pOH   =  3.04                                         pH = 10.96                                           [OH-] = 0.1 L * 0.01 M / 1.1 L                                           = 9.09 * 10-4 M                               pOH   =  3.04                                         pH = 10.96              

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