![pH=-log[H^{+}]](/tpl/images/0561/5751/863f0.png)
![[H^{+}]=e^{-pH} = \frac{1}{e^{pH} } = \frac{1}{e^{4.67} } =\frac{1}{107}](/tpl/images/0561/5751/73bd4.png)
![[H^{+}]=9.35 \times 10^{-3} M](/tpl/images/0561/5751/7ada9.png)
![K_{a} =\frac{[H^{+}][A^{-}] }{[HA]} =\frac{(9.35 \times 10^{-3}) ^{2} }{0.035}\\ \\K_{a} = 2497.78 \times 10^{-6}](/tpl/images/0561/5751/3e86c.png)
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A 0.035 M solution of a weak acid (HA) has a pH of 4.67. What is t...
Answers: 2
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