Question

The Pka of acetic acid is 1.7x10^-5. The PH of a buffer prepared by combining 50ml of 1M potassium acetate and 50ml of 1M acetic acid is:_______

Answer 1

`pH=4.77`

Step-by-step explanation:

Hello!

In this case, since the initial acetic acid and its conjugate base are mixed, we need to compute the new concentration via the final volume (100 mL):

`M_(HA)=M_(A^-)=(50mL*1M)/(100mL) =0.5M`

Which means the concentration of the acid and conjugate base are the same, therefore, the pH of the solution equals the pKa of the acid:

`pH=pKa+log(([A^-])/([HA]) )\\\\pH=-log(1.7x10^(-5))+log((0.5M)/(0.5M) )\\\\pH=4.77+0\\\\pH=4.77`

Best regards!