<p>Sorry, missed a piece of data. [OH-] in the 0.0180M solution is 5.60E-4. All solutions are at 25C and volumes are assumed to be additive. (I didn't put the state symbols in the reaction - too lazy - everything is (aq) except the H20 which is (l)</p>
<p>omg im hella freaking out about the FRQ! the MC not so much, but the FRQ will definetely kill me!</p>
<p>Is the [OH-] you gave us at equilibrium or before?</p>
<p>Actual wording:
In aqueous solution, ammonia reacts as represented above. In 0.0180M NH3(aq) at 25C, the hydroxide ion concentration, [OH]-, is 5.6E-4. In answering the following, assume that temperature is constant at 25C and that volumes are additive. </p>
<p>Implication - yes, you're at Eq.
That's the last time I try to save on typing. ;-)</p>
<p>Wow, I am stumped (as I thought I would be).</p>
<p>I was fine I think until after e part i.</p>
<p>I got
a. keq=[NH4+][OH-]/[NH3]
b. pH=10
c. Kb=1.7 x 10^-5
d. 3.1%
e, i. 30 mL HCL</p>
<p>Are those right? Can somebody please show work for e part ii and iii? Preferably with an equilibrium "box" ( with initial, delta, and eq values for all equilibrium reactants and products).</p>
<p>Answers you've given are fine except for (b). For pH, only numbers after the decimal are significant, so you've just given an answer with no sig figs. ;-) Proper answer is 10.748.
I'll work on showing e par ii and iii, but it's take me a minute</p>
<p>The proper answer should be 11 I believe since [OH-] has only 2 sig figs. Thanks for all the help and I await your explanation for the other part.</p>
<p>e part ii
(1) Find moles of acid and base
moles H+ = .0120 x .0150 = 1.8E-4
moles NH3 = .0180 x .0200 = 3.6E-4
(2) Determine what's left after neutralization reaction
After reaction -- 1.8E-4 moles of NH3 left. 1.8E-4 moles of NH4+ formed. <a href="3">Both in the same volume (.0350 L)</a> Based on species present, choose pH method
Since it's a mixture of a weak acid/weak base, use Henderson Hasselbalch. Concentrations of weak acid and base are the same, so pH = pKa = 9.255</p>
<p>Don't you mean pH=9.241?</p>
<p>Please do part iii. Thanks.</p>
<p>My bad - the problems actually said 5.60E-4 - I missed the zero when I copied it. (I hate typing all these numbers :-p). You've missed the point, though. pH=11 has NO sig figs. Numbers before the decimal in pH's don't count for sig figs. Think about it. if [H+]=1.0x10^-11, saying pH=11 is only taking information from the exponent. pH=11.00.</p>
<p>Answer key accepts both 9.255 and 9.241. Depends on whether you rounded the Kb in part c to 1.8 or 1.7</p>
<p>part iii
(1) Find moles of acid and base
moles H+ = .0120 x .0400 = 4.8E-4
moles NH3 = .0180 x .0200 = 3.6E-4
(2) Determine what's left after neutralization reaction
After reaction -- 3.6E-4 moles of NH4+ formed. 1.2E-4 moles H+ leftover (excess past the equivalence point)
(3) Based on species present, choose pH method
Since it's a strong acid mixed with a weak acid, pH only depends on concentration of strong acid. pH = -log (1.2E-4/.0600L)= 2.700</p>
<p>So if you have any amount of strong stuff left, then you only consider that for calculating pH? This is because the weak acid will disassociate very little compared to how much H+ is left to begin with.</p>
<p>Not only does the weak acid dissociate very little (because, by definition, its weak) but by LeChatelier's principle, the presence of the H+ from the strong acid will shift the weak acid equilibrium to the left - even less will dissociate than normal.</p>
<p>In any mixture of acids or bases, choose the strongest and calculate pH based on that species. (If the mixture contains more than one strong acid, they both contribute.)</p>
<p>There are four areas on a titration curve, requiring 3 methods for calculating pH. (Assuming you're titrating a weak with a strong)
1) initial pH - only weak species present - use a weak acid (or base) icebox
2) anywhere between initial and equivalence point - mixture conjugate pair - use Henderson-Hasselbalch (see e part ii)
3) at the equivalence point - only the weak conjugate of the starting species is present - use a weak base (or acid) icebox
4) Beyond the equivalence - only the leftover strong species matters - pH=-log[]</p>
<p>"Not only does the weak acid dissociate very little (because, by definition, its weak) but by LeChatelier's principle, the presence of the H+ from the strong acid will shift the weak acid equilibrium to the left - even less will dissociate than normal."</p>
<p>Wow, that just confused me.</p>
<ol>
<li>What equation are you talking about?</li>
<li>What weak acid are you talking about (NH4+ I presume)</li>
</ol>
<p>On the one hand, it's not very important. It's the justification for ignoring a weak acid in the presence of a strong acid. You really just need to know the rule. But if you care....</p>
<p>Generically, HA <--> H+ + A-
For this titration, once all the NH3 is neutralized NH4+ <--> NH3 + H+ .</p>
<p>The Ka for ammonium is small, thus little comes apart. (Equilibrium lies far to the left.) If only ammonium was present, you could do a weak acid icebox, and determine how much NH4+ dissociates. From the icebox, X=[H+] and would be low.</p>
<p>But, NH4+ isn't the only thing present. Since extra HCl was added, the H+ from the HCl affects the NH4+ equilibrium by the common ion effect.</p>
<pre><code> NH4+ <--> NH3 + H+
</code></pre>
<p>I Inital[NH4+] 0 [H+] from HCl
C -x +x +x
E [NH4+]-x x [H+]+X</p>
<p>If you solved this icebox, you would find that NH4+ dissociates less when HCl is present than when NH4+ is alone. X (the H+ from the NH4+) would be smaller, and can be ignored compared to [H+] from the HCl.</p>
<p>Clear as mud? :-)</p>
<p>Yea, I already knew the low disassociation of a weak acid in the presence of a strong acid, but what confused me was you saying to the left and not specifying an equation. Thanks.</p>
<p>any one know the format for the FR?
how many mins each q?</p>
<p>do you guys think there will be a nuclear chem frq tomorrow?</p>