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(A) According to Henry's Law,$P_{CO_2} = K_H 	imes [CO_2]$.Given that $44 \ g$ of $CO_2$ $(1 \ mol)$ in $1 \ kg$ of water corresponds to $30 \ bar$,w

Vedclass · Accepted Answer

$37$

Vedclass · Answer

(A) According to Henry's Law,$P_{CO_2} = K_H 	imes [CO_2]$.Given that $44 \ g$ of $CO_2$ $(1 \ mol)$ in $1 \ kg$ of water corresponds to $30 \ bar$,we have $30 = K_H 	imes 1$,so $K_H = 30 \ bar \cdot kg \cdot mol^{-1}$.For the soft drink at $3 \ bar$,the concentration $[CO_2] = \frac{P}{K_H} = \frac{3}{30} = 0.1 \ M$.For the dissociation $H_2CO_3 ightleftharpoons H^+ + HCO_3^-$,the concentration of $H^+$ is given by $[H^+] = \sqrt{K_{a1} 	imes C} = \sqrt{4.0 	imes 10^{-7} 	imes 0.1} = \sqrt{4.0 	imes 10^{-8}} = 2.0 	imes 10^{-4} \ M$.$pH = -\log[H^+] = -\log(2.0 	imes 10^{-4}) = 4 - \log 2 = 4 - 0.3 = 3.7$.Thus,$pH = 37 	imes 10^{-1}$.

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