# chemistry

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a 1.00 L aqueous solution contained 5.80 g of NaOH. what is the pH of the solution?
Can someone help me out with the setup of this?

• chemistry - ,

M = moles/L
moles = grams/molar mass
moles NaOH = 5.80 g/40 = moles and that will be the molarity since it is in 1 L of solution.
Then pOH = -log(OH^-) and pH + pOH = 14.

• chemistry - ,

okay, can the ph go over 14?
5.80/40=1.45
-log(1.45)= -0.16
14-^-0.16= 14.16
i must be doing something wrong

• chemistry - ,

No, you're doing it right. It's the problems that make it do this. The pH can be negative numbers and if these are pOH numbers, then pH will be larger than 14. In real world chemistry, it isn't the concn that is plugged into that formula but the activity of the solution. The activity approaches the concn at very very dilute solutions but at higher concns there are large deviations from so-called ideal behavior. Also, water has a leveling effect so that, in theory at least, 14 is the maximum number. Neither activity nor leveling effect get much discussion in beginning courses. I wouldn't let this bother me. Just go ahead and calculate using pH = -log(H^+) and pH pOH = 14.

• chemistry - ,

5.80/40 is 0.145