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November 29, 2014

November 29, 2014

Posted by **Jan** on Saturday, January 20, 2007 at 11:34pm.

[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.

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