<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)
- initial pH - only weak species present - use a weak acid (or base) icebox
- anywhere between initial and equivalence point - mixture conjugate pair - use Henderson-Hasselbalch (see e part ii)
- at the equivalence point - only the weak conjugate of the starting species is present - use a weak base (or acid) icebox
- 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>