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October 24, 2014

Homework Help: chemistry

Posted by Anonymous on Thursday, February 12, 2009 at 5:31am.

can you check if I did the problems correctly? thanks!

a soln of NaOH is prepared by dissolving 5.0g of NaOH in 500mL of water, the nadding additional water to make 1.0 L of soln. What is the pOH and pH of this soln?

[OH-]=(5g/40g)mol/1L
pOH=-log(1/8)=.90

[H3O+]=(1.0x10^-14)/(1/8)=8x10^-14
pH=-log(8x10^-14)=13.10

What is true about strong acids and bases?
*It is impossible to calculate pH from the hydroxide ion concentration
*pOH is not related to the proton concentration
*the concentration of hydroxide ions or proton is equal to the concentration of the substance dissolved
*pOH is equal to -pH

Statement 1 is wrong in saying that it is "impossible" I could also eliminate statement 4. I think either statement 2 or 3 is correct. Which one is it?

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