Asked by kevin
                A buffer is prepared using acetic acid, CH3COOH, (a weak acid) and sodium acetate, CH3COONa (which provides acetate ions, the conjugate base), according to the following proportions:
Volume of CH3COOH(aq): 131.0 mL
Concentration of CH3COOH(aq): 1.247 M
Volume of CH3COONa(aq): 118.0 mL
Concentration of CH3COONa(aq): 1.036 M
What is the concentration of the conjugate base, CH3COO-, component in the final buffer solution?
Give your answer in M.
            
        Volume of CH3COOH(aq): 131.0 mL
Concentration of CH3COOH(aq): 1.247 M
Volume of CH3COONa(aq): 118.0 mL
Concentration of CH3COONa(aq): 1.036 M
What is the concentration of the conjugate base, CH3COO-, component in the final buffer solution?
Give your answer in M.
Answers
                    Answered by
            kevin
            
    is this right?
1.247x131 /1000 = 0.16 mol
131+118=249ml
0.16/249 x 1000 = 0.656 M
    
1.247x131 /1000 = 0.16 mol
131+118=249ml
0.16/249 x 1000 = 0.656 M
                    Answered by
            DrBob222
            
    It appears to me you've calculated the concn of the acid component. The base is 118 mL of 1.036 M
    
                    Answered by
            kevin
            
    1.036x118 /1000=.12225
.122/249 x 1000 = .4910
    
.122/249 x 1000 = .4910
                    Answered by
            DrBob222
            
    That looks ok. There is a much easier way, in my opinion, to do this.
Concn is 1.036 and the initial volume of 118 is diluted to 249 so
1.036 M x 118 mL/249 mL = 0.491 M final concentration.
    
Concn is 1.036 and the initial volume of 118 is diluted to 249 so
1.036 M x 118 mL/249 mL = 0.491 M final concentration.
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