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![pH = pKa + log\frac{[base]}{[acid]} \\pH = -log(1.8 \times 10^{-5} )+log(\frac{0.202M}{0.145M} )\\pH=4.8](/tpl/images/0396/7567/f6259.png)
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Calculate the ph of a buffer that is 0.145 m hc2h3o2 and 0.202 m kc2h3o2. the ka for hc2h3o2 is 1.8...
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