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The partial pressure of $\ce{CO2}$ gas in a bottle of carbonated water is $3.84\mathrm{~atm}$ at $25\mathrm{~^\circ C}$. How much $\ce{CO2}$ gas (in g) will be released from $0.5\mathrm{~L}$ of the carbonated water when the partial pressure of $\ce{CO2}$ is lowered to $1.04\mathrm{~atm}$? $25\mathrm{~^\circ C}$, the Henry’s law constant for $\ce{CO2}$ dissolved in water is $1.65 \times 10^3\mathrm{~atm}$, and the density of water is $1.0\mathrm{~g/cm^3}$.

First, I found the number of moles of $\ce{CO2}$ using this formula:

$$n = \frac{PV}{k}$$

So the answer that I found which is $0.000351$, can I multiply it by a molar mass of a $\ce{CO2}$ to get the number of grams inside the bottle? This approach seems too easy for me, but at the same time it makes sense.

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  • $\begingroup$ @Ghost nice editing, but you missed out one CO2 :) Also, we normally add the units inside math mode, you can check out the current code to see how it's done. $\endgroup$ Oct 31, 2015 at 14:34

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This shows part of the problem.

Henry's law as given is

$K_{AW}^{px} = $1.65*10^3$ = \frac{p_{\ce{CO2(gas)atm}}}{[\ce{CO2(aq)}]}$

where: $K_{AW}^{px}$ atm = aqueous-phase mixing ratio of $\ce{CO2}$ and the units are $atm$

$[\ce{CO2{(aq)}}]$ = the concentration of $\ce{CO2{(aq)}}$ in $mole/L$

$p_{\ce{CO2(gas)atm}}]$ = the partial pressure of $\ce{CO2{(gas)}}$ in $atm$

So rearranging...

$[\ce{CO2(aq)}] =\frac{p_{\ce{CO2(gas)atm}}}{K_{AW}}$

The moles of $\ce{CO2}$, $m_{\ce{CO2}}$, is given by concentration, $[\ce{CO2}]$, times the volume, $v$ in $L$.

$m_{\ce{CO2(aq)}} = [\ce{CO2(aq)}]v = \frac{p_{\ce{CO2(gas)atm}}}{K_{AW}}v$

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  • $\begingroup$ It does not seem right to me, if you use the formula Hpv and substitute all values in it which will be: 1650*0.5*1.04 = 858 grams. This number does not make sense to me, how can 858 grams of CO2 be in 0.5 L of bottle? $\endgroup$
    – student123
    Oct 31, 2015 at 15:23
  • $\begingroup$ Could you double check the constant? Is it $1.65*10^{-3}$. $\endgroup$
    – MaxW
    Oct 31, 2015 at 15:28
  • $\begingroup$ No, it says it is 1.65*10^3 $\endgroup$
    – student123
    Oct 31, 2015 at 15:32
  • $\begingroup$ This table en.wikipedia.org/wiki/… shows the problem. You didn't give any formula for the Henry's law and there were a number of them. I'm sorry. Instead of guessing one, I should have clarified that point. I'll fix the analysis. $\endgroup$
    – MaxW
    Oct 31, 2015 at 15:51
  • $\begingroup$ In such cases a dimensional analysis is good to be sure that the right units are being used. $\endgroup$
    – MaxW
    Oct 31, 2015 at 16:50

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