# Calculating the pH of CO2 dissolved in 1l of water

 P: 16 OK folks, first of all, let me thank those of you who have helped me with previous questions.... perhaps this is why I keep coming back! This time, I'm just asking those of you who are better than I am at this if I did this right and thus, got the correct result. I think I did, but since this is not a homework problem, I have no way of checking my answer... I set up my problem like this (see the attached diagram): Imagine our system has a beaker with 1 liter of water and is enclosed in a constant-pressure (of 1 atm.) atmosphere of 100% carbon dioxide at 25°C. We let the carbon dioxide come to equilibrium with the water so the maximum dissolves. What would the resultant pH be? What do we know?the pKa of carbonic acid is 6.367 the solubility of carbon dioxide in water at 25ºC 1atm is 1.45 gram/liter we have 1 liter of water What is the reaction? CO2 + 2H2O → H2CO3 + H2O → HCO3- + H3O+the number of moles of carbon dioxide that dissolve in water at equilibrium will equal the number of moles of carbonic acid in solution. 1.45 g. CO2$\frac{1 mol.}{44.01 g.}$= 0.032947 mol. CO2Calculate using a RICE table (see the attached image table): Convert pKa to Ka: 10-pKa = Ka → 10-6.367 = KaKa (10-6.367) = $\frac{[A-][H+]}{[HA]}$ = $\frac{(x)(x)}{(0.0329-x)}$can we apply the "rule of 500?":Rule of 500: if $\frac{[HA]}{Ka}$>500, ignore any nonzero changes in "x"$\frac{0.0329}{1x E-6.367}$=76,594 >> 500 so yes.simplify Ka:10-6.367=$\frac{x²}{0.0329}$x²=10-6.367(0.0329)=1.41317x10-8x=√1.41317x10-8=1.18877x10-4=[H+]pH = -log[H+]=-log(1.18877x10-4)=3.9249≈3.92so the lowest the pH can go in 1 liter of water in equilibrium with 100% CO2 at 25ºC and 1 atmosphere pressure is 3.92. It sounds like a reasonable number, or did I totally mess something up somewhere? thanks so much for any help on this! Attached Thumbnails