Final Exam Blog Response

pH and pOH:

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

_{2}?*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_{2}

0.0200 x 2 = 0.0400 M

^{-}, 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

^{+})**Looking for pOH**

given information:

V

_{1}= 50.0 mL = .0500LpH = 1.007

V

_{2}= 100.0 mL = .100 LYou will end up using the equation M

_{1}V_{1}= M_{2}V_{2}First, you need to find the molarity of M

_{1 }(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 M

_{1}V_{1}= M_{2}V_{2}, you can solve for M_{2}V

_{1}= 50.0 mL = .0500L V_{2}= 100.0 mL = .100 L M

_{1}= .0984 M M_{2}= ?M

_{1}V_{1}= M_{2}V_{2}M

_{2}= M_{1}V_{1}/V_{2}M

_{2}= (.0984 X .0500)/.1000M

_{2}= .0492 M 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**
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.

ReplyDeleteThanks!

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.

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

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!

ReplyDelete