~~~~~~~~~~~~~~~~~~~~~~~~~~~~~CHEM202~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Quiz /29/ min

Transcription

1 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~CHEM202~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Quiz /29/ min a) ph= -log(1.8e-5) + log(0.1/0.1) ph = 4.74 b) H + + CH 3COO - _ CH 3COOH complete limiting reagent problem to get mol CH 3COO- and mol CH 3COOH ph= -log(1.8e-5) + log(0.025/.075) = 4.27 (note I have used no of moles instead of molarity in log(base/acid). This is allowed as long as the two species are in the same volume of solution, which is usually the case.) 2. Find Limiting Regent: 2g / g/mol = moles of Zinc 0.1 L * 0.50 mol/l = moles of HCl this is the Limiting Reagent due to the 2 to 1 ratio moles * 0.5 = moles of H 2 produced. V = nrt / P = (0.025)*( )*(298)/(1) = L of H 2 3. a ) We cannot use ph=pka+log(base/acid) because there is no base, which means we would have to compute log(0), which would yield a nonsensical ph. We must, therefore, treat this as a regular equilibrium problem for the dissociation of the acid CH3COOH (acetic acid). CH 3COOH _ CH 3COO - + H + We would then have to solve Ka = x 2 /(0.2-x), giving x= =[h+] ph = -log[h+] = 2.72 b)ch 3COOH + OH - _ CH 3COO - Finish limiting reagent problem to get mol CH 3COO - and mol CH 3COOH

2 ph = pka + log (base/acid) ph = 4.74 log (.005/.015) ph = 4.27 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~CHEM202~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Quiz /29/ min a) ph= -log(6.8e-4) + log(0.3/0.3) ph = 3.17 b) HF + OH- _ F- Complete the limiting reagent problem to get 0mol HF, 0.28mol OH- and 0.24mol F-. We cannot use ph = pka + log(base/acid), because we have no more acid left! We would have to evaluate log(.24/0), which would yield a nonsensical ph. Therefore, we must consider the species left in solution, and set up an equilibrium problem for F-. F- _ HF + OH- Converting back to concentrations, we have initial concentrations of [F-] =.24/( ) =0.4M and [OH-] =.28/(.4+.2)= 0.47M. After completing the ICE table, we end up having to solve: Kb = (Kw/Ka) = x(.47+x)/(0.4-x). Which gives x = 1.73e-11. This means [OH-] ~ 0.47M. Thus [H+] = Kw/[OH-]= 2.12e-14. ph = -log[h+] = 13.67

3 2. PV=nRT n=mass/molar Mass PV=(mass/Molar Mass)RT P*Molar Mass = (mass/v)rt P*Molar Mass = density*rt Molar Mass = (density*rt)/p =39.99 g/mol Argon 3. HClO 4 is a strong acid. Therefore, we will not be using ph = pka + log(base/acid) as the pka is undefined and nearly equal to zero. Instead, in a) c), since there is acid left in solution, using ph = -log[hclo 4] = -log[h+], we can find ph. In part d, all the acid is consumed by the base, and there is no more of either species left in solution. Since all the salts formed do not change ph (because they do not have conjugate acids/bases of weak bases/acids), the ph will be neutral. a. 0 ml base added ph = b. 10 ml c. 40 ml d. 80 ml 7.00 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~CHEM202~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ Quiz /29/ min a) Kb = 1.8e-5. Ka = Kw/Kb = 5.56e-10 ph= -log(5.6e-10) + log(0.4/0.4) ph = 9.25 b) NH 3 + H + _ NH 4 + complete limiting reagent problem to get mol NH3 and mol NH4+ ph= -log(5.6e-10) + log(0.085/.135) = 9.05 (note I have used no of moles instead of molarity in log(base/acid). This is allowed as long as the two species are in the same volume of solution, which is usually the case.)

4 2. Firstly, let us assume for simplicity, that the total volume is 1L. Then, using the information given and PV=nRT, we can solve for n, which is the total number of moles of methane AND helium. Plugging P=1atm, V=1L, R=.08206LatmK-1mol-1, we get n= mol. We also know the mass of helium + methane mixture = g, because we have the density of the mixture (helium + methane), and we have assumed a volume of 1L, and mass = density x volume. Now let us assume that of this g, x grams of helium is present. This implies that ( x)g of methane is present. Now, using no of moles = (mass/molar mass), lets consider: No of moles(total) = No of moles(helium) + No of moles(methane) mol = (x/4) + (( x)/16) (mass/molar mass) This gives x = g = mass of helium Thus, no of moles(helium) = x/4 = 3.817e-4 mol. Now, volume % of helium = (volume of helium/volume of helium+methane) x 100 Since V = (RT/P)n, we can just as well say Volume % helium = (no of moles (helium)/no of moles(total))x100 = (3.817e-4/.0446)x100 = 0.86% (Note answers will vary from 0.86% to roughly 1.35% depending on if you rounded values and used approximate molar masses. Without rounding, and using accurate molar masses, you should get about 1.35%) 3. H2NNH2 is a weak base. a) Treat like a normal equilibrium problem for the reaction of a weak base with water. H2NNH2 _ H2NNH3 + + OH - Initial concentration = 0.1 M. Complete the ICE table, and the resulting equation to solve is:

5 Kb = x 2 /(0.1-x) x = 5.47e-4 = [OH-] [H+] = Kw/[OH-] = 1.83e-11 ph = -log[h+] = c) HNO 3 is a strong acid. H 2NNH 2 + H + _ H 2NNH 3 + Complete the limiting reagent problem to get.006 mol H2NNH2 and.004 mol H2NNH3+. We can now use ph = pka + log(base/acid) Ka = Kw/Kb = 3.33e-9 ph = -log(3.33e-9) + log(.006/.004) ph = 8.65