Bootcamp Chem Question

[drdmn]

I am having trouble with a question on bootcamp, I would appreciate any explanation! Here is the question:

What is the pH of a solution prepared by adding 20 mL of 0.02 M NaOH(aq) to 20 mL of water?

I get that we have to use M1V1=M2V2, but H2O has 2 H, while NaOH has 1 OH. I though that when this is the case, we have to use normality, so (1)(0.02)(0.02) = M2(0.04)(2).

Am I going about this the wrong way? Is it different because water is involved?

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[Carbohydrates]

H2O has 2H but remeber that it dissociated in to H+ and OH-, not 2H+..

As for the problem, you have the dilution equation right, so calculated the concentration of H+ in the final solution and figure out the pH from there. Also you can reasoning about this problem, water has a pH of 7, then when you add NaOh which is a base, then the pH gonna be larger then 7..

 

[deleted696582]

Water only has one H to donate to a solution. Remember that it dissociates into OH and H? Anyway, you don't have to use M1V1 = M2V2.

Molarity = mol/L, so 0.02 M = mol/0.02L, which gives you 0.0004 mol NaOH. This means you're adding 0.0004 mol of NaOH to 20 mL of water. It's a dilution.

The final volume of the solution is 40 mL, and you've added 0.0004 mol NaOH, so the Molarity = 0.0004 mol / 0.04 L, which equals 0.02 M.

0.02 M is equal to 2 pOH (not pH, because base has been added), so the pH should be 12.

 

[Futuredentista62]

All you have to do is figure out the molarity of NaOH in 40 ml because that the new volume.
You're adding 20 more ml so now you have 4o ml.

M1V1=M2V2
(.02)(.02)=M(.04)

M=1X10-2
So concentration of NaOH= 1x10-2, and [OH-]=1x10-2
POH=2
PH= 14 - POH= 12

So, PH=12

 

[drdmn]

Thanks all, I understand now!