Express the following in mol/kg
[H+]in a water with a pH of 6.8?
Ok, so the pH=-log10[H+]. Here we have a pH of 6.8 so we have...
6.8=-10log10[H+]
solving we get...
[H+]=10^-6.8 mol/L
converting from Liters to m^3 and dividing by the density of water we get...
(10^-6.8 mol/L)*(L/0.001m^3)*(m^3/1000kg)=1.585(10^-7)mol/kg
Hope that helps!
Thanks.
I have a few more problems...can you help?
how about expressing 230 parts per thousand (ppt) of SO4^2- in mol
per kg?
You are likely to have more luck if you repost as a new question.