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:

·      pH = -log(H⁺)

·      (H⁺) =10 ^–pH

·      pH = 14 – pOH

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

ANSWERS:

a)    pH = 12.602

                  b)    (OH^-) = 1.7 X 10^-11 M

   c)    pOH = 12.692