Sunday, May 15, 2011

Final Exam Review #27

Final Exam Blog Response

pH and pOH:

 a)  What is the pH and pOH of a 0.0200M solution of Ca(OH)2?

b) What is the concentration of hydrogen and hydroxide ions in a solution of HCL with a pH of 3.24?

 c) A 50.0 mL solution contains has a pH of 1.007 before being diluted to a volume of 100.0 mL . What is the pOH of the final solution?

Useful equations pertaining to pH and pOH:

·      pH = -log(H+)
·      (H+) =10 –pH
·      pH = 14 – pOH

·      pOH = -log(OH-)
·      (OH-) = 10 –pOH
·      pOH = 14 – pH

-       Sig Figs: for pH, they only count after the decimal place
-       (H+) / (OH-) = molarity

 a) 0.0200 M of Ca(OH)2

- Ca(OH)2 is a base, so you are going to use pOH

You have to multiply the molarity by two since it is (OH)2
            0.0200 x 2 = 0.0400 M

Since you have the molarity of OH-, and you are looking for the pOH, use:
     pOH = -log(OH-)
     pOH = -log(0.0400)
     pOH = 1.398

*sig figs: using regular sig figs rules, the molarity has three sig figs. When finding the pH or pOH you only count sig figs after the decimal place. Therefore, the 1 in 1.398 doesn’t count as a significant figure, so you have to have three sig figs after the decimal, (.398)

     It is asking for the pH, and since you know the pOH you use:
     pH = 14 – pOH
     pH = 14 – 1.398
     pH = 12.602

b)  pH = 3.24; HCl; looking for (H+) and (OH-)

HCl is an acid, so you are going to use pH.

Since you are only given the pH, you have to use:
        (H+) =10 –pH
                        (H+) =10 -3.24
                        (H+) = 5.8 X 10-4 M

*sig figs: pH has two numbers after the decimal, therefore it has two sig figs. You do not count the three because it is before the decimal point. Since there is a total of two sig figs for pH, there is a total of two sig figs for (H+). You use the regular rule sig figs rule, so 5.8 = two sig figs.

You then need to find (OH-). Since you need the pOH first in order to find the (OH-), you use:
pOH = 14 – pH
pOH = 14 – 3.24
pOH = 10.76

Now that you know the pOH you can find the (OH-)          
            (OH-) = 10 –pOH
            (OH-) = 10 –10.76
            (OH-) = 1.7 X 10-11 M

*sig figs: same rule applies to this as it applied to find the (H+)

 c) Looking for pOH 
given information:
V1 = 50.0 mL = .0500L
pH = 1.007
V2 = 100.0 mL = .100 L

You will end up using the equation M1V1 = M2V2

First, you need to find the molarity of M1 (H+)
                (H+) =10 –pH
(H+) =10 -1.007
        (H+) = .0984 M

*sig figs: pH has three significant figures after decimal, so molarity will have a total of three sig figs

Now that you have all the necessary information for M1V1 = M2V2, you can solve for M2
V1 = 50.0 mL = .0500L            V2 = 100.0 mL = .100 L
                                    M1 = .0984 M                           M2 = ?

M1V1 = M2V2
M2 = M1V1/V2
M2 = (.0984 X .0500)/.1000
M2 = .0492 M
                                (H+) = .0492

             Now you have the (H+) for the final solution. Solve for the pH of this solution:
pH = -log(H+)
pH = -log(.0492)
pH = 1.308

Last step: solve for the pOH:
pOH = 14 – pH
pOH = 14 – 1.308
pOH = 12.692

a)    pH = 12.602
                  b)    (OH-) = 1.7 X 10-11 M
   c)    pOH = 12.692


  1. Christina, you did a great job explaining your thought process and guiding me through the steps in your problem. As far as I can see all your problems and sig figs are done correctly.

  2. I disagree with your value for [OH-] on part B. I think it should be 1.7x 10^-11, which I found using Kw. The answer you provided does not check, but you method is correct in my opinion. I think you may have mistyped the numbers into the calculator.

    Other than that, great job. Your explanations were clear and thorough.

  3. Thanks Sahil! You're right about the [OH-] value. I just reentered it into the calculator and 1.7 X 10^-11 is the correct answer. I will be sure to fix that!