Abuffer is prepared by adding 20.0 g of sodium acetate 1ch3coona2 to 500 ml of a 0.150 m acetic acid 1ch3cooh2 solution. (a) determine the ph of the buffer. (b) write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the

First, we have to get moles of CH3COONa = mass/molar mass

 = 20 g / 82.03 g/mol = 0.244 moles 

when we have  [CH3COOH] = 0.15 M

∴ [CH3COONa] = moles of CH3COONa / Volume of solution
                           
                           = 0.244 moles / 0.5L = 0.488 M

when we look up for Ka of acetic acid value it is equal 1.8 x 10^-5

So we can get Pka = -㏒Ka
                             
                                = -㏒(1.8 x10^-5)
                                = 4.7 

now we will use Henderson - Hasselbalchn equation to get the PH:

PH = Pka + ㏒[conjugate basic/weak acid]

when CH3COOH is the weak acid & CH3COO- is the conjugate base so by substitution:

PH = 4.7 + ㏒ (0.488/0.15)

      = 5.2

b) when we have this equation for the reaction:

HCl + CH3COONa → CH3COOH + NaCl

ionic equation : H+ + Cl- + CH3COO- + Na+ → CH3COOH + Na+ + Cl-

when HCl + H2O → H3O+  + Cl-

∴ the reaction will be:

CH3COO- (aq) + H3O+(aq) → CH3COOH(aq)  + H2O(l)