Question

The pH of buffer solution prepared by mixing 50mL of 0.2M CH3COOH and 25mL of CH3COONa is 4.8 what is the concentration of CH3COONa? Pka of CH3COOH=4.8.

Answer 1

The concentration of [CH3COONa] will be 0.133M

concentration of [CH3COOH],

we have 50 ml of CH3COOH = 0.050 L.

Hence,

[CH3COOH] = 0.050*0.2/ 0.075 = 0.133 M

We know that,

pH = pKa + log[CH3COO-]/[CH3COOH]

=> 4.8 = 4.8 + log[CH3COO-]/[CH3COOH]

=> log[CH3COO-]/[CH3COOH]  = 0

=> [CH3COO-]/[CH3COOH] = 1

=> [CH3COO-] = [CH3COOH] = 0.133 M

Hence the concentration of  CH3COONa is 0.133M.

Answer 2

Answer:

Formula:-

ph=pka+log[v×CH3COONa]/[v×CH3COOH]

Process:-

4.8=4.8+log[25×CH3COONa]/[50×0.2]

log[2.5×CH3COONa]=0

2.5×CH3COONa=1

CH3COONa=1/2.5

Final Answer Is

therefore concentration of CH3COONa is 0.4molar

I hope this answer may help you