[Steve1956]

It's been a long time since I've done this kind of chemistry, so I want to make sure I'm doing this problem correctly.

Question: You plan to add CaO to increase the pH of a lake.  The current ph is 4 and you would like to raise the ph to 6.5, the volume of your lake is 4.0 x 109 L. How much CaO do you need to add? 1 mole of CaO neutralizes 2 moles of H+

My Work:

I know the reaction is CaO+H₂O ::equil::Ca(OH)₂

Start of water: pH=4, pOH=10.
                     [OH-]= 10^-pOH
                             = 1x10^-10 M OH

End Desired water: pH=6.5 pOH=7.5
                          =3.16227766x10^-8 M OH

Diff of 3.1522776610^-8 M OH

So, to calculate the amount of CaO I solved for X here (multiplying by 2 because 1 mol of CaO produces two moles of OH-)

(X moles of CaO/4x10⁹ L H₂O)*2=3.16227766x10^-8 M OH

x=63.04555 moles of CaO needed

then 63.04555 moles of CaO * 56 grams/mol= 3530.5508 g CaO needed to increase the pH.

[Borek]

No, you need to remove H⁺, that means CaO reacts with H⁺ producing Ca²⁺ and water.

Your reaction can be seen as a first step in the process - note that OH^- will immediately react with H⁺.

[Steve1956]

I don't get what you're saying here. So what do I need to do?

[Borek]

Write reaction equation between CaO and H⁺, calculate how much H⁺ has to be consumed, calculate how much CaO you need for that.

[Steve1956]

Ok, so the reaction then is CaO+2H⁺   Ca²⁺ + H₂O

Now I'm blanking on how to figure out how much H+ has to be consumed.

[Steve1956]

ok, so maybe now I know what i'm doing, but as I said its been a long time since i've had gen chem for these problems.

To calculate moles of H+ in the volume of water given:

Start: 10^-4*4 * 10⁹= 400000 moles H+

Want: 10^-6.5*4 * 10⁹= 1264.9111 moles H+

so we take the difference to find out how many moles we need to neutralize:
400000-1264.9111=398735.08894 moles of H+ to neutralize

since 1 mol CaO neutralizes 2 H+, we divide that number by 2 to give 199367.544468 moles of CaO needed

199367.544468 moles of CaO* 56= 11164582.4902 grams of CaO needed?

[Borek]

Correct approach, I guess reporting it as 11.2 (or even 11) tonnes will do.

[Steve1956]

Borek, thank you very much for your help, I appreciate it!