Practice Test 1 Solutions - CHEM 112 Exam 3 1E This is a straight up solubility question with no real complications. The equation and ICE table we set up is: AgI 2 (s) Ag 2+ (aq) + 2I - (aq) I - 0 0 C - +s +2s E - s 2s Notice that we don t have to worry about AgI 2 (s) since it s a solid and isn t relevant to the equilibrium calculations. Next we set up the equilibrium equation: K sp = [Ag 2+ ][I - ] 2 = (s)(2s) 2 = 4s 3, where s is the molar solubility. Then we solve for x:! s = K $ sp # &13 = 0.0015 " 4 % Lastly, we must recognize that the question asks for [I - ] which according the ICE table is equal to 2s = 2(0.0015) = 0.0030 M.
2D This is a strong acid / strong base titration. To begin, we must compare the number of moles of acid with the number of moles of base and see which one is our limiting reactant: Moles HClO 4 = 0.15 L x 0.1 M = 0.015 mols Moles OH - = 0.1 L x 0.5 M = 0.05 mols Since there are fewer moles of acid they will all be consumed by the base and we will be left with: 0.05 0.015 = 0.035 mols OH -. All we need do now is calculate [OH - ] and then convert to ph: [OH - ] = (0.035 mol) / (0.25 L) = 0.14 M (note that the volume is the combined volumes of both the acidic and basic solutions) poh = -log(0.14) = 0.85 ph = 14 poh = 14 0.85 = 13.15 3E This is a surprisingly straightforward question that seems to cause a lot of confusion. We know that the dissolution of Ag 2 CO 3 looks like: Ag 2 CO 3 (s) 2Ag + (aq) + CO 2-3 (aq) And so at equilibrium we write: K sp = [Ag + ] 2 [CO 2-3 ], which we then use to solve for [Ag + ]: [ Ag + ] = 8.1 10 0.025 12 = 1.8 10 5 M
4B To get E 0 cell we need to find the half reactions that are taking place and look up their standard electrode potential values: Al(s) Al 3+ (aq) + 3e -, E 1/2 = -1.66 V Sn 2+ (aq) + 2e - Sn(s), E 1/2 = -0.136 V Keep in mind that the top reaction involving Al will appear in the reverse direction as a reduction but we don t have to change the sign of its half-cell potential if we use the equation: E 0 cell = E 0 cathode - E 0 anode = (-0.136) (-1.66) = 1.524 V Remember that reduction occurs at the cathode and oxidation at the anode which is why the half-cell values go where they do, but even if you messed up and got them in the wrong order you d still get the same number just as a negative instead of a positive. If that happens you can just take the absolute value since most questions will indicate that this is a voltaic or galvanic cell, which means that E 0 cell must be positive. However, this question makes no mention as to whether or not the electrochemical cell is indeed voltaic, so technically you have to make sure that yo have the right reaction for each half cell before you subtract the values.
5B This a standard weak acid / strong base titration. First we must calculate the number of moles of weak acid and the number of moles of base: Moles HNO 2 = 0.1 L x 0.5 M = 0.05 mols Moles OH - = 0.1 L x 0.05 M = 0.005 mols Now we should look at the reaction that takes place and set up an ICE table: HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O(l) I 0.05 0.005 0 - C -0.005-0.005 +0.005 - E 0.045 0 0.005 - As the table suggests, since we re dealing with a strong base it reacts completely, hence at equilibrium there is no hydroxide left and all we ve got in solution is some of the weak acid we started with, HNO 2, and some of its conjugate base, NO 2 -. The Henderson Hasselback Equation is perfect for situations like this since it s a direct link to ph when all you ve got in solution is weak acid / conjugate base. So we plug in our values: ph = -log(4.5 x 10-4 ) + log(0.005/0.045) = 2.4 (note that when using the Henderson Hasselback equation it s not necessary to convert our equilibrium mole values to concentrations)
6A Looking at the information provided, it would seem that the best way to solve for ΔG would be to use: ΔG 0 = -nfe 0. Since F is just a constant, all we need is to find E 0 and n. We know that the reduction reaction with the most positive half-cell potential is the reaction that actually is the reduction reaction and that the other reaction must be the oxidation reaction. From that we can solve for E 0 by plugging in the values: E 0 = E 0 cath - E an = (+1.50) (-0.14) = +1.64 V The positive value confirms that we have the right order since we re told this is a voltaic cell. Next we have to calculate n. This requires us to balance the two reactions so that the number of electrons lost equals the number of electrons gained. The two reactions are: Sn(s) Sn 2+ (aq) + 2e - Au 3+ (aq) + 3e - Au(s) The only way to balance these efficiently is multiply the first reaction by 3 and the second by 2: 3Sn(s) 3Sn 2+ (aq) + 6e - 2Au 3+ (aq) + 6e - 2Au(s) Thus we see that number of electrons transferred is actually 6, and that is our n value. Finally we plug in everything to get: ΔG 0 = -nfe 0 = -(6)(9.65 x 10 4 )(1.64) = -949560 J = -950 kj 7A The primary requirement of any buffer system is the presence of a weak acid and its conjugate base. Since (a) is the only option that has a weak acid (HF) and its conjugate base (F - ) that is the correct answer. Remember that RbF in solution becomes Rb + and F - since Rb is a group 1 metal.
8D First we need to separate the reaction into two half reactions, and then apply material and charge balances to each: MnO 4 - (aq) + 1e - MnO 4 2- (aq) I - (aq) + 3H 2 O(l) IO 3 - (aq) + 6H + (aq) + 6e - Once this is done we must multiply each reaction by the smallest whole number to get the electrons transferred to be equal. This means that we just multiply the first reaction by 6 and leave the other one alone to get: 6MnO 4 - (aq) + 6e - 6MnO 4 2- (aq) I - (aq) + 3H 2 O(l) IO 3 - (aq) + 6H + (aq) + 6e - Remember that when balancing we add an H 2 O for every oxygen needed and an H + for every hydrogen needed. Now we combine the reactions again by adding together everything on the left side and everything on the right side (make sure to cancel out anything that exists on both sides): 6MnO 4 - (aq) + I - (aq) + 3H 2 O(l) 6MnO 4 2- (aq) + IO 3 - (aq) + 6H + (aq) Note that this would represent the balanced reaction in acidic solution. The next step is to take into consideration that we re in basic solution and so we must add an OH - ion to each side of the reaction for every H + that s present to get: 6MnO 4 - (aq) + I - (aq) + 3H 2 O(l) + 6OH - (aq) 6MnO 4 2- (aq) + IO 3 - (aq) + 6H + (aq) + 6OH - (aq) As expected the 6H + (aq) and 6OH - (aq) will neutralize to form 6H 2 O and the reaction becomes: 6MnO 4 - (aq) + I - (aq) + 3H 2 O(l) + 6OH - (aq) 6MnO 4 2- (aq) + IO 3 - (aq) + 6H 2 O(l) Lastly, we cancel out any extra H 2 O that might be on both sides to get: 6MnO 4 - (aq) + I - (aq) + 6OH - (aq) 6MnO 4 2- (aq) + IO 3 - (aq) + 3H 2 O(l) This is our balanced reaction in basic solution. The sum of the coefficients is 23.
9B Since we re not told if this is a voltaic cell or an electrolytic cell, it s safer to make sure that we know where each half reaction happens: 2H + (aq) + 2e - H 2 (g), E 0 1/2 = 0.00 V This is clearly a reduction and therefore occurs at the cathode. Pb(s) Pb 2+ (aq) + 2e -, E 0 1/2 = -0.126 V Now we can calculate E 0 cell: This is clearly an oxidation and therefore occurs at the anode. E 0 = E 0 cath - E an = (0.00) (-0.126) = +0.126 V Because E 0 cell is positive we now know that this is a voltaic cell and that ΔG 0 should be negative, which means we can eliminate answers (D) and (E). Lastly we plug in our values into: ΔG 0 = -nfe 0 cell = -(2)(9.65 x 10 4 )(0.126) = -24318 J/mol = -24 kj/mol 10C Remember that the salt bridge is designed to provide ions (not electrons) to both half cells so as to maintain charge neutrality. Thus (C) is the answer. Keep in mind that these ions are usually non-reactive and certainly do not react with anything, especially at the anode/cathode.
11B There are a couple of ways to solve this. One is time consuming, the other is not. Under most circumstances you would need to calculate both the number of moles of weak acid, ammonium (NH 4 + ), and that of the weak base, ammonia (NH 3 ), so that they could be used in the Henderson Hasselback equation. This is the time-consuming way. Alternatively, you can hopefully recognize that both the acid and base solutions have the same volumes and concentrations and so their concentrations in the buffer are equal to one another: [NH 4 + ] = [NH 3 ] This is referred to as the halfway point, and that means that ph = pk a. So all we really need to do is determine pk a : 14 K w 10 K a = = = 5.6 10 5 K 1.77 10 b ph = pk a = -log(5.6 x 10-10 ) = 9.25 10 12A The oxidizing agent is the species that gets reduced. This can only be a reactant so (d) and (e) aren t even options to begin with. The quickest way to answer this problem is to assign oxidation numbers to each element on both the product s side and the reactant s side and see which elements show a change in number. If this is done correctly you will notice that the Cr in Cr 2 O 7 2- has an oxidation number of +6 on the reactant s side and +3 on the product s side. This means that it since its charge is decreasing it must be gaining electrons and so it is being reduced. Thus, Cr 2 O 7 2- must also be the oxidizing agent. As a side note, I d like to point out that the oxidation number for S in S 2 O 3 2- is +2 but on the product s side it is +2.5. This is very unusual, and yet we still see an increase in oxidation number, which means that S 2 O 3 2- is still being oxidized and is thus the reducing agent.
13B If we set up the reaction for the solubility product constant, K sp, we get: Mg(OH) 2 (s) Mg 2+ (aq) + 2OH - (aq), And we can write: K sp = [Mg 2+ ][OH - ] 2 hydroxide = [x][2x] 2, where x is the solubility of magnesium K sp = 4x 3 = 4 x (1.4 x 10-4 ) 3 = 1.1 x 10-11 14D The question is really asking us to find out the molar solubility of Mn(OH) 2 in a solution that already has hydroxide ions floating around. This is the common ion effect, and means that the initial concentration of hydroxide has to be taken into account when calculating solubility at equilibrium. Since ph = 11.5, we know poh = 14 11.5 = 2.5, and so [OH - ] = 10-2.5 = 0.0032 M To help calculate solubility we ll use the following ICE table: Mn(OH) 2 (s) Mn 2+ (aq) + 2OH - (aq) I - 0 0.0032 C - +x +2x E - x 0.0032 +2x We then set up the equation for the solubility product constant, K sp : K sp = [Mn 2+ ][OH - ] 2 = [x][0.0032 + 2x] 2, where x is the solubility Since K sp is so small we can assume that x is negligible and the equation becomes: K sp = [x][0.0032] 2 x = (1.6 x 10-13 ) / (0.0032) 2 = 1.6 x 10-8 M By examining the ICE table we can see that [Mn 2+ ] = x, and so [Mn 2+ ] = 1.6 x 10-8 M
15C For something to spontaneously oxidize something else, we need to look for the best oxidizing agent which is above (and to the left) the reaction dealing with Br -. Remember that the best oxidizing agents are always at the top left of the cell potential chart. Of the available options, only Cl 2 (g) is above our bromide oxidation.
16B Since Cr(NO 3 ) 3 readily dissolves (recall solubility of NO 3 - ions) we really have a solution with 0.01 moles of Cr 3+ and, as the solution is basic, some hydroxide ions. The question clearly suggests these ions are going to react to form Cr(OH) 4 -, based on the reaction given. More importantly we must recognize the extremely high K f value that goes with this reaction. Typically we ve been setting up equilibrium calculations that deal with reactions that have relatively low equilibrium constants (like ten raised to some negative power). In this reaction our equilibrium constant, K f, is 8 x 10 29, which basically means that the reaction will go completely towards product formation until our reactant is almost all used up. I say almost, because a tiny, small amount will still remain, and that amount is the answer to our question. To calculate [Cr 3+ ] we need to set up the equilibrium constant equation, but first we need to get the concentrations of [OH - ] and [Cr(OH) 4 - ]. To calculate the initial [OH - ]: poh = 14 ph = 4 [OH - ] = 10-4 Keep in mind that because the question explains that the solution is buffered at ph = 10, the [OH - ] isn t going to change much at all during the reaction and so we can assume that the hydroxide concentration remains constant. Since the balanced reaction shows a 1:1 ratio between Cr + and Cr(OH) 4 -, we can assume that the 0.01 moles of Cr 3+ will form 0.01 moles of Cr(OH) 4 -, and so [Cr(OH) 4 - ] at equilibrium will be (0.01 mol) / (1 L) = 0.01 M. Technically the concentration should be the tiniest bit lower since not all of the chromate ions were used up, but the difference is so small it is negligible. Lastly we set up the equilibrium equation: K f [ Cr 3+ 4 [ Cr( OH ) = 3+ [ Cr ][ OH [ Cr( OH ) ] = [ K ][ OH f ] ] 4 4 ] 4 ] = ( 0.01) 29 4 ( 8 10 )( 10 ) 4 = 1.25 10 16
17B This is about as difficult a question you ll likely see regarding electrolysis. To determine what M is, we ll probably have to figure out its molar mass and then see which answer choice has the same value. Molar mass = (mass / moles), and since we already have the mass, we just need to know how many moles of the metal, M(s), are going to be formed during the electrolysis. Since the metal oxide is M 2 O 3, we know that the metal must have a +3 charge and will undergo the following reaction: M 3+ + 3e - M(s). Now let s use the information provided: - 7.5C 1mol e 1mol M(s) 7.5A = 1800s 4 - s 9.65 10 C 3 mol e = 0.0466 mol M(s) So Molar mass = 1.26g 0.0466 mol = 27 g / mol Of the answer choices available, the only one with a molar mass close to 27g/mol is Al.
18E Since the cell potential, E 0, is positive, we know we re dealing with a voltaic cell and thus we can predict which reactions are taking place. This is important because the equation we ll need to solve this question requires us to know which species are reactants and which are products. However, the reduction of Ag 2+ to Ag(s) is not given on your data sheet and so technically this problem can t be solved for sure without that piece of information (sorry, my bad). Nonetheless, were the data sheet to have that reaction, the reductions would be: Ag 2+ (aq) + 2e - Ag(s), E 1/2 = -0.746 Pb 2+ (aq) + 2e - Pb(s), E 1/2 = -0.126 Since the reduction of Pb 2+ is more positive than that of Ag 2+, we know that Pb 2+ is reduced and its half reaction stays as is, while the Ag half reaction must be flipped since it has to now be an oxidation. Thus the overall redox reaction is: Ag(s) + Pb 2+ (aq) Pb(s) + Ag 2+ (aq) Thus we use the following equation to solve for [Ag 2+ ]: E ce ll = E 0 ce ll 0.059V logq n 2+ 0.059 [ Ag ] 0.65 = 0.62 log 2 [0.35] Note that we use n = 2 as there are 2 electrons transferred during the balanced reaction. We can now solve for [Ag 2+ ] = 0.038 M
19A This is another equation dealing with Faraday s law, except this time we need to solve for current. Remember that the units of current are Amps = C/s, so really we re looking for Coulombs and time (in seconds). Well, time is easy, since we can just convert 25 minutes to 1500 seconds. Next we have to consider the reaction that has to take place in order to end up with Pt(s), so we must first look at the solution it comes from, PtCl 4 (aq), which tells us that the charge on the Pt ion must be +4 and so the reaction (reduction) that must occur is: Pt 4+ (aq) + 4e - Pt(s) Now we use the information provided to determine the charge needed: - 4 1mol Pt 4 mol e 9.65 10 C 15g - 195 g 1mol Pt 1mol e = 29692 C All we do now is plug in our values for charge and time to get current: current = C s 29692C = 1500s = 19.8 A 20D Another electrolysis problem. This time we re solving for mass, which almost always means solving for number of moles and then converting to mass. Before we begin calculations, make sure you understand that the 0.50M concentration is irrelevant for solving this problem, and is just designed to throw you off. So firstly, we should look at the type of solution we re dealing with to determine the reaction (reduction) that must take place. Copper(II) sulfate means we have a Cu 2+ ion in solution and so the reaction we are looking for is: Cu 2+ (aq) + 2e - Cu(s) Now we ll use the information given to get the moles of Cu(s) produced: - 2.3C 1mol e 1mol Cu(s) 63.5 g.3a = 1500s = 1. 14g 4 s 9.65 10 C 2 mol e 1mol 2 -
21E Simply use the Henderson-Hasselbalch equation because it is a buffer system of a weak acid and its conjugate base. The concentration of the acid is 0.15 M and 0.23 M for the base. ph = pka + log [base]/[acid] 22E Below the belt of stability we are neutron deficient, and so a proton must be converted to a neutron to increase stability. This can be achieved by either positron emission or electron capture.
Practice Test 2 Solutions CHEM 112 Exam 3 1B. An atom in its elemental form always has an oxidation state of zero; therefore, Al(s) has an oxidation state of zero. Al(s) loses electrons (is oxidized) when it becomes Al(OH) 4 -, because it then has an oxidation state of +3. A substance that is oxidized is a reducing agent. 2D. To balance a redox reaction, first separate it into the reduction and oxidation half reactions. Oxidation: Mn2+ MnO4- Reduction: NaBiO3 Bi3+ Balance each of these separately. First balance all atoms other than O and H (for example add Na+ to the right side of the oxidation to balance the Na atoms). Then, balance the O atoms, by adding the same number of H2O molecules to the other side. Then, balance the H atoms (including those from the H2O s just added) by adding H+ ions to the other side. Finally, balance the charge on each side by adding electrons to either the reactant side (reduction equation), or the product side (oxidation equation). The final step is to add both equations together, making sure that the number of electrons on the reactant side and product side are equal. For this to be true in this equation, the oxidation equation had to be multiplied by 2 and the reduction equation had to be multiplied by 5. Therefore, the final coefficient in front of Bi3+ is five. 3E When balancing a redox reaction in basic solution you use the same approach as in acidic solution until the end when everything else has already been balanced. Then, you add OHions to each side for every H+ present. In this case, there should be 14H+ on the reactant side, so we add 14 OH- to both sides. After the H+ and OH- on the reactant side neutralize to form H2O, you will then have 14 H2O on the reactant side and 7H2O on the product side, which cancel out to leave 7 H2O on the reactant side.
4D This is a weak acid / strong base titration. As with similar titrations we must first calculate the number of moles of acid base: Moles HNO 2 = 0.1 L x 0.25 M = 0.025 mols Moles OH - = 0.05 L x 0.5 M = 0.025 mols Immediately, you should recognize that the initial number of moles of weak acid are equal to the number of moles of hydroxide added, and so we are at the equivalence point. This means that all of the weak acid and hydroxide will be used up and the only thing we ll have left in solution that can alter the ph is the conjugate base, NO 2 -. Furthermore, we know that we have the exact same number of moles of NO 2 - (0.025 mols) as we had of the hydroxide ions that were all used up during the titration. This is all useful information if you have any interest in saving time during your test. Still, for those who want to see a more detailed analysis, the ICE table below shows what s going on: HNO 2 (aq) + OH - (aq) NO 2 - (aq) + H 2 O(l) I 0.025 0.025 0 - C -0.025-0.025 +0.025 - E 0 0 0.025 - As you can see, all we have left in solution is 0.025 moles of NO 2 -, and so to calculate ph we just have to treat this problem as we would any other weak base equilibrium problem by using the equation: K b = x M 2 i Remember that since we re dealing with a weak base we ll be needing K b (not K a ) and also that x = [OH - ] (not [H 3 O + ]). We get K b using the same old equation: 14 Kw 10 K b = = = 2.2 10 4 K 4.5 10 a 11
Next we need the initial molarity, M i, which is the concentration of our weak base: [NO - 2 ] = (0.025 mol) / (0.15 L) = 0.17 M (note that the volume is the combined volumes of both the acidic and basic solutions) Now we can solve for x: K b x = x = M K 2 b i, M i = 11 ( 2.2 10 ) ( 0.17) = 1.9 10 6 = [ OH ] poh = log[ OH ph = 14 poh = 8.3 ] = log 1.9 6 ( 10 ) = 5.7 5C. A common ion (F-) exists in solution. Therefore, the solubility of PbF 2 will decrease. The Ksp expression in terms of solubility looks like: Ksp = (s)(0.5 +2s)^2 Do not forget to square the concentration of F- in the expression. Plug in the value of Ksp given and solve for s. You can neglect the +2s to make for an easier calculation. The value for s (molar solubility) is your answer. 6B. The see which salt will precipitate first as NaI is added, you must figure out the solubility of each salt given based on the initial concentration of ions. For example, using the Ksp for AgI, the equation to find AgI s solubility is as follows: 8.3 x 10^(-17) = [(3.3 x 10^(-3) + s) * s] The 3.3 x 10^(-3) represents the initial concentration of Ag+. When you solve for s you get an extremely small number. One that is smaller than the solubility of PbI 2. Therefore, AgI will precipitate first.
7B. The best reducing agents can be easily identified on the Standard Reduction Potential Chart in the book or on the exam data sheet. They are the products listed on the bottom of the chart; the lower on the chart the better the reducing agent. Since Zn (s) is the PRODUCT lowest on the chart, then it is the best reducing agent. 8A. The oxidation state of Ag+ is 1+ and it is reduced to Ag(s), which has an oxidation state of zero. When the oxidation state of an ion/molecule is lowered, that ion/molecule is REDUCED. 9A. As electrons move from anode to cathode, a difference in charge builds up at each compartment. The reaction will eventually stop if the cathode becomes too negatively charged. The anions and cations migrate to the anode and cathode, respectively, to neutralize this charge build up and keep the reaction going. 10D At the equivalence of a weak acid titration statement ii is correct. Statement i is true at the half-way point. Also, at the equivalence point, the only thing left in solution that could affect ph is the conjugate base of the weak acid, and so since it s a base we cab expect the solution to be basic too. Thus ph > 7. Note that if this were a STRONG acid / strong base titration, then the ph would equal 7 at the equivalence point.
11B For optimum buffer capacity we first need to find weak acid / conjugate base pairs. Answer choices (a) and (c) both use HCl (a strong acid) and cannot be candidates for the best buffer. Since (b) and (d) both use the same legitimate weak acid / conjugate base pairs we must then look to see which pairs have the highest minimum concentrations. The minimum concentration for answer (d) is 0.15M, whereas for answer (b) it is 0.25M, so the most reliable of the two buffers is (b) as it has the highest minimum concentration. 12D. The standard cell potential of this redox reaction can be found using the table on your data sheet. E(cathode)-E(anode) gives a positive value. Therefore, the reaction is spontaneous, so K>>>1 and G is negative. 13E. The last statement is true. This process is known as electroplating. 14B. Salts that break up into ions that will act as a weak or strong base will have increased solubility in acidic solution. AgCl does not have this characteristic because Cl- is the conjugate of a very strong acid.
15A This is a strong acid / strong base titration. We fist need to determine the number of moles of hydronium and of hydroxide: H 3 O + : 0.05M x 0.13L = 0.0065 moles OH - : (0.05M x 0.055L) x 2 mol OH / 1 mol Ba(OH) 2 = 0.0055 moles If we set up an ICE table for this reaction we ll see that all the hydroxide is used up and we are left with 0.0010 moles of H 3 O +. Thus the ph can now be determined by calculating the final concentration of hydronium and then converting to PH: [H 3 O + ] = (0.0010 moles / 0.185 L) = 0.0054 M, and ph = -log[h 3 O + ] = 2.27. 16A. The reduction process takes place at the cathode. This will involved adding electrons to the cation in the solution to make the metal form. 17C. When a battery dies, its redox reaction stops because the cell has reaction an equilibrium state. At equilibrium, the cell potential is ZERO. 18A. The metal least likely to corrode is the one that would be least likely to be OXIDIZED. Therefore, the metal HIGHEST on the standard reduction potential list will be the most likely to be REDUCED, and therefore best at avoiding being oxidized relative to the other metals below it on the standard reduction potential list.
19D. Base this question off the formula: Ecell = Ecell(standard) (0.059/n)*logQ If you increase the concentration or pressure of a product in a reaction, it will drive the reaction back into the formation of the reactants. Driving a reaction back to the left will DECREASE the cell potential of a redox reaction. Note that solids are NEVER included in the Keq/Q expressions, so their amounts will not affect Ecell. Hydrogen gas is a product and it exists in the Q expression for this reaction, so increasing its value will decrease Ecell. 20E. The E cell cannot be found using the standard reduction table because both reactants are not listed. However, Keq is given, so E cell can be found using the formula E cell = (0.059/n) * log K Plug in for K and 1 for n because only one electron is involved in this redox reaction as shown in the change in oxidation states. E cell comes out to 0.229V. However, we want Ecell because we are NOT at standard conditions. Q = [Fe3+][VO2+] / [Fe2+][VO 2 +][H+]^2 Ecell = Ecell(standard) (0.059/n)*logQ All the values will cancel except for the [H+], which you find by taking the inverse log of the ph given. Therefore, Q = 1 / (1 x 10^(-6))^2. Plug this value into the above equation along with the calculated Ecell(standard), to get a final answer of -0.479.
21B. The main objective will be to determine the ph (and then the poh) which will require the number of moles of acetic acid and NaOH present. That calculation is as follows: (0.05 L)(0.2 M) = 0.01 moles acetic acid (0.035 L)(0.1 M) = 0.0035 moles NaOH Since there are less moles of base, it is the limiting reagent in the reaction with the acid and will run out first. Therefore, 0.01 0.0035 gives 0.0065 moles of acidic acid formed. The 0.0035 moles of OH- from NaOH led to the production of 0.0035 moles of acetic acid s conjugate base (the acetate ion). The total volume in this solution is 0.085 L, which the number of moles of acetic acid and its conjugate base must each be divided into to get their concentration values. Now, plug those values into the Henderson- Hasselbalch equation to determine ph, subtract that value from 14, and you get your poh. 22C. The buffer zone exists between ph = 4 and ph = 6, and this curve shows a weak acid being titrated with a strong base.