Question

a one molar solution of HCl is about 92 percent ionized at room temperature .Calculate Ka for HCl.....(give me only solution no details.)

Answer 1

Answer on Question #56580 - Chemistry - Inorganic Chemistry

Question:

A one molar solution of HCl is about 92 percent ionized at room temperature. Calculate Ka for HCl...(give me only solution no details.)

Answer:

$$\mathrm{HCl} = \mathrm{H}^{+} + \mathrm{Cl}^{-}$$ $$Ka = \frac{[H^{+}] [Cl^{-}]}{[HCl]}$$ $$[H^{+}] = [Cl^{-}]$$ $$\mathrm{CHCl} = [\mathrm{HCl}] + [\mathrm{Cl}^{-}]$$ $$x = [\mathrm{Cl}^{-}] / \mathrm{CHCl}; [\mathrm{Cl}^{-}] = x^* \mathrm{CHCl} = 92 \%*1 \mathrm{M} = 0.92 \mathrm{M}$$ $$[\mathrm{HCl}] = 1 \mathrm{M} - 0.92 \mathrm{M} = 0.08 \mathrm{M}$$ $$Ka = \frac{[H^{+}] [Cl^{-}]}{[HCl]} = \frac{0.92 \mathrm{M} * 0.92 \mathrm{M}}{0.08 \mathrm{M}} = 10.58 \approx 11$$

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