Answer:
[tex]pH=4.77[/tex]
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):
[tex]M_{HA}=M_{A^-}=\frac{50mL*1M}{100mL} =0.5M[/tex]
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:
[tex]pH=pKa+log(\frac{[A^-]}{[HA]} )\\\\pH=-log(1.7x10^{-5})+log(\frac{0.5M}{0.5M} )\\\\pH=4.77+0\\\\pH=4.77[/tex]
Best regards!
