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I'm studying for my final exam, and these are the questions I got wrong on the mid term, could anyone help correct me? I know people are against giving answers for homework, so I must add that is now exam period for Universities, so this is only for my further understanding! Thanks so much!


My attempts will be in this format


1)The pH of a 0.2M unknown base X OH is 8.15

a)Calculate [OH-]


10^(-8.15) = 7E-9M


b)What is the % disassociation


(7E-9/0.2M) x 100 = 3.5E-6%


c)What is the Kb of the base?


This one I don't even know where to begin..


2)Calculate the pH of the following solutions

a) 0.10M KCN, Ka (HCN) = 6.2E-10 (Hint: KCN is the salt of a weak acid and a strong base)


x = sqrt(Ka*c) x = sqrt(6.2E-10 * 0.1) = 7.8E-6 -log(7.8E-6) = pH = 5.1


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closed as not a real question by ManishEarth Apr 6 '13 at 6:46

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. If this question can be reworded to fit the rules in the help center, please edit the question.

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    $\begingroup$ Welcome to Chemistry.SE! For this type of question, we would like you to help us help you. What progress have you made on these problems? What don't you understand? It's possible your wrong answer is a simple arithmetic mistake instead of a conceptual problem. As it stands, we don't like questions that seem to suggest we should do your work for you. $\endgroup$ – Ben Norris Apr 5 '13 at 19:27
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    $\begingroup$ In addition to what @BenNorris said, in the future, please ask only one question per post. (You may ask more questions if they are closely related, though). You've already gotten an answer, but if you want this reopened, just tweak it so that you ask one question and show what you have tried, and I'll reopen it :) $\endgroup$ – ManishEarth Apr 6 '13 at 6:48
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1a) First use $pOH = 14 - pH$

Then use: $pOH = \log([OH^-])$

b) Simply use: $\dfrac{\text{dissociated}\ [OH^-]}{\text{original}\ [OH^-]}$

c) Use $K_b = \dfrac{[X^+][OH^-]}{[XOH]}$

2) Dissolving $KCN$ in water gives: $\ce{KCN \rightarrow K^+ + CN^-}$

The following equilibrium reaction is thus formed

$\ce{CN^- + H2O \rightleftharpoons HCN + OH^-}$

Which is:

$K_b = \dfrac{[HCN][OH^-]}{[CN^-]}$

From the $K_a$, you can get the $K_b$ through $K_w\ =\ K_aK_b$ so that:

$\dfrac{K_w}{K_a} = \dfrac{[HCN][OH^-]}{[CN^-]}$

Let the amount of $CN^-$ that reacts to be $x\ M$, thus $[CN^-]$ becomes $0.1-x$ and you get $x\ M$ of both $[HCN]$ and $[OH^-]$. Solve for $x$ to get $[OH^-]$ and from there, $pH$ should be easy to calculate.

3) Use $Ksp = [Ba^{2+}][F^-]^2 = 1.7\times 10^{-6}$ and solve for $[Ba^{2+}]$.

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  • $\begingroup$ Thank you so much! To clarify, my error on the first question was I used pH instead of pOH. So I did 10^-8.15, when I should have done 10^(- (14-8.15) )? $\endgroup$ – BrettD Apr 5 '13 at 19:46
  • $\begingroup$ @BrettD Yes, since by taking the pH, you are actually finding the number of moles of $[H^+]$. You could still continue if you used the other equilibrium expression: $K_w = [H^+][OH^-] = 1.0\times10^{-14}$. $\endgroup$ – Jerry Apr 6 '13 at 12:05
  • $\begingroup$ for 2) should be $6.2\times10^{-10} = \dfrac{[0.1-x]^2}{[x]}$ $\endgroup$ – DavePhD Dec 16 '14 at 13:38
  • $\begingroup$ @DavePhD Oh my, that was long ago hmm... I'll opt for another way of doing that. Updated. Also had a typo in there. $\endgroup$ – Jerry Dec 16 '14 at 19:43

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