The AP Chemistry Study Thread

Test Preparation AP Test Preparation
161

<p>smder, as the titration curve approaches the equivalence point, the concentration of OH-/H+ is decreasing.</p>

<p>think of it this way.</p>

<p>say you start w/ HCl w/ a Ph of 0 (1.0M HCl) and like 100mL of it.
you start adding NaOH..... you add 10mL of 1.0M NaOH</p>

<p>titration curves usually have Ph on the y-axis, volume of NaOH (in this case) on the x-axis.
what is the Ph after the 10mL of NaOH? .1moles HCl- .01moles NaOH
= .09moles HCl into 110mL of NaOH</p>

<p>what is the molarity of that-------- .09/.110 = .8181M
-log(.8181) = .08715 which is very close to 0. so adding 10mL of NaOH to 100mL of HCl barely changes the Ph.</p>

<p>go up to 100mL 1.0M HCl w/ 50mL 1.0M NaOH
.1mol HCl + .05mol NaOH --> .05 mol HCl left
150mL of solution
.05/.150 =.333M
Ph of that = .47756 --> still not even at 1. </p>

<p>now go to 80mL
.1mol HCl + .08mol NaOH --> .02mol HCl left.
180mL
.02/.180=.11111111M
Ph of that = .9542--> STILL not even at 1!
80mL of NaOH and not even at 1!</p>

<p>that's why it isn't linear......... if you continued at this rate using just up to 80mL at extrapolating to future Ph, it wouldn't go above like 2 or 3.</p>

<p>let's look at 95mL NaOH now.
.005 mol HCl left
195mL--> M = .0256M
ph of that = 1.5911--> finally above 1, but still very low (remember that 7 is neutral), that's only 5mL away from equivalence point!</p>

<p>look at 99mL NaOH.
.001 mol HCl left
199mL -->.005025M
Ph = 2.298864--> not even 7, and this is ONE mL away from equivalence point!</p>

<p>at 100mL, Ph = 7..... think of the jump between that! take the slop assuming it's linear.
7-2.298864/(.1-.099) = 4.7011/.001 = 4701.1--> slope of the line.... very VERY high.</p>

<p>that's why there's such a straight line straight up near Ph.</p>

162

<p>what colors do we have to know. ex: calcium=orange, sodium-yellow, etc</p>

163

<p>The following reaction is nonspontaneous at 25C:</p>

<p>Cu2O(s) --> 2Cu(s) + .5 O2(g) change in G = 141 kJ</p>

<p>If change in S (entropy) is 75.8 J/K, above what temperature will the reaction become spontaneous?</p>

164

<p>You guys will have to look up numbers yourselves.</p>

165

<p>bumppity bump</p>

166

<p>HC3H5O2 <--> C3H5O2 + H+ K= 1.34 x 10^-5</p>

<p>1.) Propanoic Acid ionizes in water in the according to the equation above.</p>

<p>a.) Write equilibrium-constant expression for the reaction.</p>

<p>b.) Calcualte pH of .265 M solution of propanoic acid.</p>

<p>c.) .496g NaC3H5O2 is added to 50 mL sample of 0.265 M Solution of propanoic acid.
i.) Concentration of C3H5O2
ii.) Concentration of H+</p>

<p>HCO2 + H2O <--> HCO2H + OH-</p>

<p>d.) Given that [OH] is 4.18 x 10^-6 in .309 M solution of Sodium Methanoate, calculate the following:
i.) Value of Kb for HCO2
ii.) Value of Ka for HCO2H</p>

<p>e.) Which acid is stronger, propanoic acid or methanoic acid? Justify your answer</p>

167

<p>1a)</p>

<p>keq = ([C3H5O2][H+]) / [HC3H5O2]</p>

<p>1b)
__HC3H5O2 <--> C3H5O2 + H+
I---.265 M---------0 M----0 M
C--- -x ----------- +x --- +x
E--- .265-x ------- x ----- x</p>

<p>keq = ([C3H5O2][H+]) / [HC3H5O2]
1.34 x 10^-5 = x^2 / (.265-x)
Assume x is small...so [HC3H5O2] = .265 (we'll check this later)</p>

<p>1.34 x 10^-5 = x^2 / (.265)
(1.34 x 10^-5)(.265) = x^2
3.55*10^-6 = x^2
.0019 = x</p>

<p>x is a small number...so the assumption is alright to make (.265-x = .265)</p>

<p>x represents [H+]</p>

<p>pH = -log[H+]
pH = -log[.0019]
pH = 2.72</p>

<p>1c) (.496 g)(1 mol / 96 g) = 0.0052 mol NaC3H5O2</p>

<p>assuming that NaC3H5O2 takes up no addl vol</p>

<p>[NaC3H5O2] = (.0052 mol / 50 mL)(1000 mL / 1 L) = 0.104 M
NaC3H5O2 dissociates into Na+ and C3H5O2-</p>

<p>therefore [C3H5O2-] = 0.104 M (before rxn)</p>

<p>__HC3H5O2 <--> C3H5O2 + H+
I---.265 M-------.104 M----0 M
C--- -x ----------- +x --- +x
E--- .265-x ----- .104+x ----- x</p>

<p>1.34 x 10^-5 = [(x)(.104+x)] / (.265-x)
1.34 x 10^-5 = (.104x + x^2) / (.265-x)
(.265-x)(1.34 x 10^-5) = (.104x + x^2)
3.55<em>10^-6 - 1.34</em>10^-5x = (.104x + x^2) </p>

<p>-x^2 - 1.39x + (3.55*10^-6) = 0
quadratic equation</p>

<p>x= </p>

<p><something went="" wrong="" with="" 1c...gonna="" look="" at="" it="" when="" im="" done="" the="" rest="" of="" prob=""></something></p>

<p>EDIT: I probably made a math mistake somewhere....just read the reply by evilboya</p>

168

<p>I already replied to him via PM.
here is the message I sent him. not too many explanations, i'll explain anything you guys don't understand.</p>

<p>1a)
Ka = [C3H5O2-][H+]/[HC3H5O2]</p>

<p>b)
1.34E-5 = x^2/.265
x^2 = 3.551E-6
x = .001884409 = [H+]
Ph = -log[H+]
Ph = 2.7248</p>

<p>c)
.496g NaC3H5O2 = .0052 mol NaC3H5O2
.0052/.05L = .103M NaC3H5O2
NaC3H5O2 --> Na+ + C3H5O2-
.103M NaC3H5O2 = .103M C3H5O2-</p>

<p>HC3H5O2<-->H+ + C3H5O2-</p>

<p>1.34E-5 = .103x/.265
x = 3.4476E-5 = [H+] = additional [C3H5O2-]
i. [C3H5O2-] = .103 + 3.45E-5 = .103M
ii. [H+] = 3.45E-5M</p>

<p>d)
HCO2- + H2O <--> HCO2H + OH-
NaHCO2 --> Na+ + HCO2-
.309M .309M
[OH-] = 4.18E-6
Kb = [HCO2H][OH-]/[HCO2-]
[HCO2H]=[OH-]
i. Kb = 4.18E-6^2/.309
Kb = 5.6545E-11</p>

<p>ii. Kb(Ka) = 1E-14
1E-14/Kb = Ka
Ka = 1.77E-4</p>

<p>e)
Ka propanoic = 1.34E-5
Ka methanoic = 1.77E-4</p>

<p>Methanoic is stronger because the Ka is larger, meaning that it dissociates more in water than propanoic does</p>

169

<p>1e) the larger the ka for an acid...the stronger it is (the more it dissociates)</p>

<p>ka for methonoic (formic acid) is 10^-3.75 (google it up)
for proponic 1.34 x 10^-5</p>

<p>therefore formic acid is stronger</p>

170

<p>Wow....evilboyya....you got to it before I did.</p>

171

<p>G = H - TS
G + TS = H
141000J + (298K)(75.8J/K) = H
H = 163.59kJ.</p>

<p>set G = 0
G = H-TS
0 = H - TS
TS = H
H/S = T
163588.4/75.8 = T
T = 2158K</p>

<p>high on life, i don't think qualitative analysis is a significant part of the actual AP Chem test...... but I would assume basic flame test colors and ion colors would be useful if it is? perhaps someone else can answer this better.</p>

<p><em>EDIT</em>
scareya- yea I worked on it before, then saw he had also posted on this thread.</p>

<p>ctrl + C and Ctrl + V</p>

172

<p>. At 10 degrees Celcius, 8.9 x 10^-5 grams of AgCl dissolves in 100 mL of water.</p>

<p>i.) Write teh equation for the dissasociation of AgCl.
ii.) Calculate the solubility of AgCl in water
iii.) Calculate the value of Ksp for AgCl</p>

<p>b.) At 25 degrees Celcius, the value of Ksp for PbCl2 is 1.6x10^-5 and Ksp for AgCl is 1.8x10^-10</p>

<p>i) If 60mL of .04 M NaCl is added to 60 mL of .03 M Pb(NO3)2, will a precipitate form?</p>

<p>ii) Calculate the equilibrium value of Pb2 in 1 Liter of saturated PbCl2 solution to .25 mole NaCl.</p>

<p>iii.) If .1 M NaCl is added to beaker containing both .12 M AgNO3 and .15 M Pb(NO3)2 at 25 degree celcius, which will precipate first?</p>

173

<p>a)
i. AgCl <---> Ag+ + Cl-
ii. 8.9E-5g AgCl = 6.21E-7 mol in 100ml
6.21E-7/.1 = 6.21E-6mol/L
iii. Ksp = [Ag+][Cl-]
Ksp = 6.21E-6^2
Ksp = 3.856E-11</p>

<p>b)
i. 2NaCl + Pb(NO3)2 --> PbCl2 + 2NaNO3
60mL, .04M 60mL,.03M
.0024mol---.0018 mol---- 0 mol ---0 mol
.0012 mol --.0018 mol---- 0------- 0 ---
0 mol ----- .0006 mol ----.0012-- .0024</p>

<p>.0012 mol PbCl2 in 120mL of solution
.0012/.12 = .01M PbCl2 soln</p>

<p>1.6E-5 = x^2
x = .004M Pb+2 </p>

<p>precipitate will form b/c PbCl2 can only dissolve up to .004M Pb+2..... but there is .01M PbCl2, meaning there will be a precipitate.</p>

<p>someone else can finish b) ii) and b iii...... i gotta study for a test</p>

174

<p>i ddint really understand what you did in B.</p>

<p>anyone else</p>

175

<p>ok, this is what I did.</p>

<p>found the # of moles from Molarity and volume for NaCl and Pb(NO3)2
- that's first line of table</p>

<p>-second line of table is dividing by the coefficient b/c there are 2 NaCl--> let's you see which one is limiting reactant.</p>

<p>-3rd line of table
final moles of all species</p>

<p>60mL + 60mL --> 120mL solution
assume that it doesn't precipitate --> how concentrated is the solution</p>

<p>.0012 mol in 120mL --> .01M PbCl2</p>

<p>the calculations below that are to calculate the molar solubility of PbCl2..... I actually did that wrong.</p>

<p>PbCl2 <----> Pb+2 + 2Cl-
Ksp = [Pb+2][Cl-]^2
1.6E-5 = (x)(2x)^2
1.6E-5 = 4x^3
4E-6 = x^3
x = .006 = M = [Pb+2]</p>

<p>the molar solubility of [Pb+2] is less than the molarity of the PbCl2 solution, therefore, part of the solution precipitates out.</p>

176

<p>bump bump bump</p>

177

<p>if anyone has an electronic or hard copy of the 2002 ap chem test IM me at spydertennis so we can talk, i can make it worth your while</p>

178

<p>Anyone can discuss with me about 2003 AP chem free response. Or if you want it, I can send you the questions typed up. pm me (I have no sn).</p>

179

<p>Question to everyone:</p>

<p>What is Boltzmann's constant? It's on the AP formula sheet, but not in my textbook or the PR book.</p>

180

<p>Alright...I figured that question out.</p>

<p>My friend told me that the equilibrium frq isn't going to be acid-base b/c last year had an acid-base. Anybody heard of this theory?</p>

<p>Next question:</p>

<p>What's the difference between alkaline and alkaline earth metal salts from transition metal salts? (Physical appearance)</p>