Question

If a buffer solution is .110 M in a weak acid (Ka = 2.7*10^-5) and .530 M in its conjugate base, what is the pH?

Answer 1

\[pH = p Ka + \log \left[ A-/ HA \right]\]
\[pKA = - \log KA = - \log (2.7\times10^{5}) = 5.431\]
\[pH =5.431 + \log \left( 0.110/0.530\right)\]
\[pH = 5.431 -0.68 = 4.751\]