Chemistry, 13.07.2020 16:01 johnmarston60
A 5.22 × 10−3−mol sample of HY is dissolved in enough H2O to form 0.088 L of solution. If the pH of the solution is 2.37, what is the Ka of HY?
Answers: 1
![pH = -log [H^{+} ]\\\[ [H^{+} ] = antilog -pH = antilog -2.37 = 4.27 \times 10^{-3} M](/tpl/images/0703/6539/30f06.png)
![Ka = \frac{[H^{+}]^{2} }{Ca} = \frac{(4.27 \times 10^{-3} )^{2} }{0.0593} = 3.07 \times 10^{-4}](/tpl/images/0703/6539/f3e32.png)
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A 5.22 × 10−3−mol sample of HY is dissolved in enough H2O to form 0.088 L of solution. If the pH of...
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