Calculate the pH of 1.00 L of the buffer 0.96 MCH 3 Na/ 1.10 CH 3 COOH before an

Question

Calculate the pH of 1.00 L of the buffer 0.96 MCH3Na/ 1.10 CH3COOH before and after theaddition of the following species. (Assume there is no change involume) (a) pH of starting buffer: number=_________________ (b) pH after addition of 0.070 mol NaOH number=__________________ (c) pH after further addition of 0.126 mol HCl number=___________________ *********WILL RATE*********** Calculate the pH of 1.00 L of the buffer 0.96 MCH3Na/ 1.10 CH3COOH before and after theaddition of the following species. (Assume there is no change involume) (a) pH of starting buffer: number=_________________ (b) pH after addition of 0.070 mol NaOH number=__________________ (c) pH after further addition of 0.126 mol HCl number=___________________ *********WILL RATE***********

Explanation / Answer

Formula:                   pH = pKa + log [salt / acid] a) pKa for CH3COOH is 4.7447 Upon substituting the data in the equation ,                         pH = 4.7447 + log ( 0.96 / 1.10 )                                = 4.68 b ) when a strong base is added to the acidic buffer number ofmoles of salt increases and number of moles of aciddecreases. Number of moles of salt = 0.96 moles + 0.070 moles                                       =1.03 moles Number of moles of acid = 1.10 moles - 0.070 moles                                       = 1.03 As both are equal pH = pKa ,                     pH    = 4.7447 c ) when a strong acid is added to the acidic buffernumber of moles of salt decreases and number of moles of acidincreases. Number of moles of salt = 0.96 moles - 0.070moles                                       = 0.89moles Number of moles of acid = 1.10 moles + 0.070 moles                                       = 1.17                               pH  = 4.7447 + log ( 0.89 / 1.17 )                                      = 4.62 Number of moles of salt = 0.96 moles - 0.070moles                                       = 0.89moles Number of moles of acid = 1.10 moles + 0.070 moles                                       = 1.17                               pH  = 4.7447 + log ( 0.89 / 1.17 )                                      = 4.62

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