​When equal volumes of 2 M CH₃COONa and 2 M HCl are mixed, pH is found to be x and on mixing equal volumes of 2 M CH₃COOH and 2 M CH₃COONa, pH is found to be y. Hence is

1. For solution 1 in which 2 M of CH₃COONa + 2M of HCl :

 (a) pH of HCl = -log[H⁺] = -log[2] = -0.3 (highly acidic)

(b) pH of CH₃COONa (salt of weak acid and strong base)
                  = $\frac{p{K}_w+p{K}_a+\log a}{2}$                                                     (you need to provide the value of K_a and a= 2M)
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2. For solution 2 M of CH₃COOH + 2 M CH₃COONa,which is a buffer of weak acid and its salt 
                      =  pK_a + $\log \left(\frac{moles of salt}{moles of acid}\right)$                                    (again value of K_a)

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