Solubility of CO2 in Water

Q.

Solubility Product for CO2(g)<-> CO2(aq) at 25 degrees celcius is calculated to be K = 5.23x10^-2. If the the CO2(aq) in a lake at the same temperature is 2.2mg/L, is the lake in equilibrium with CO2(g)(partial pressure = 10^-3.5? If not in equilibrium, is the gas volatizing from the lake or dissolving into it? CALCULATE Q AND COMPARE IT TO K=5.23X10^-2 TO DETERMINE EQUILIBRIUM. PLEASE SHOW ME HOW TO CALCULATE Q with CO2(g)<->CO(aq). Given: CO2(aq)=2.2mg/L CO2(g) partial pressure =10^-3.5
A.
2.2mg CO2/L = 0.0022g CO2/L
[0.0022g CO2/L]/(44g CO2/mol) = 5.0x10^-5 mol CO2/L
K = pCO2 / [CO2] = 5.23x10^-2
If you substitute NON equilibrium concentrations into the equilibrium constant expression, the value you get is called Q in order to distinguish it from the equilibrium value of K.
Q = pCO2 / [CO2] when pCO2 and/or [CO2] are not at equilibrium
Q = (5.0x10^-5 mol CO2/L) / 10^-3.5 = 0.16
Since Q is larger than K, 0.16 > 5.23x10^-2, the equilibrium,
CO2(g)<-> CO2(aq)
will shift to the right
CO2 gas will dissolve to form more CO2(aq)

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