Review For Final Exam Aleks solutions PDF

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How to do every ALEKS topic in the pie. Step by step...

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Calculating absolute entropy using the Boltzmann hypothesis

S=Kb(LnW) S = entropy of system Kb = Boltzmann Constant 1.38065×10−23 W = number of microstates of the system Plug in value into Kb and W and multiply J/K units in Aleks Understanding the effect of pH on the solubility of ionic compounds Ask yourself how changing the pH (H+) concentration will affect the dissolution equilibrium of each compound.

if one of the product ions acts as a Brønsted-Lowry base then it will react with any excess H+ and be removed from solution Using Ksp to calculate the solubility of a compound - How to find solubility Ksp=[x][y] xy=aqueous ions remaining (do not use liquid or solids) See Ksp data in aleks or provided reference (see if given in question) Set up ICE table to help organize Ex. 1 Ksp=[Ba^2+][SO4^2-] Initial

0

0

Change

x

x

Equilibrium

x

x

Substituted equation = X*X Ksp=x^2 Use equation by plugging in Ksp value and solve for x. Multiply x value by formula mass of substance = solubility of compound Units in aleks = g/L Ex. 2 Ksp=[Ni^2+][OH-]^2 Difference from Ex. 1 to Ex. 2 is [y] has exponent due to balanced equation having 2 in front of OHInitial

0

0

Change

x

2x

Equilibrium

x

2x

Substituted equation = X*(2X)^2 Ksp=4x^3

Rearrange and solve for x. Plug in ksp and solve. Multiply x value by formula mass of substance = solubility of compound Units in aleks = g/L Using the solubility of a compound to calculate Ksp Ksp=[x][y] xy=aqueous ions remaining (do not use liquid or solids) Divide g/L by formula mass of substance solubility/molar mass Take previous answer and apply it to all [x] and [y] = a Concentrations with coefficients will multiply their coefficient to your previous answer (not all will have coefficients). = b [a][b]^2 = Ksp. Squared value comes from previously used coefficient. Not all equations will have. Calculating the pH of a weak acid titrated with a strong base

●

●

Strong bases react with acids to completion, so the acid, the added base, or both will be completely consumed as soon as the base is added to the acid. The solution stoichiometry problem is to decide which happens, and to calculate the molarity of the important acids or bases left in solution. If the acid is used up, you're left with an aqueous solution of strong base. If the added base is used up, you're left with an aqueous solution of a weak acid and its conjugate base. If both weak acid and strong base are used up, you're left with a solution of the acid's conjugate base. You can calculate the pH of any of these solutions using techniques you've already mastered. A= Weak Acid B= Strong base

A. Multiply molarity by volume of weak acid. Ex. .3300M * .1927L = A (mols of weak acid A) Make sure all volumes are in liters B. Multiply molarity by volume of strong base. Ex. 1.000M * .0266100L = B (mols of strong base B) Subtract A-B and see which remains. A-B=C C= remaining substance (in this case acid) Because there is more mols of acid than base remaining the mols of acid remaining can now be calculated Go down if base > acid Find molarity of unreacted acid by: C/total volume in L = Molarity of remaining acid

Mols of anion produced = OH- added so to find molarity of OH- in solution: Take previous B and do B/total volume in L B/total volume in L = molarity of OH- in solution

Need to now solve for Ka Ka=[H+][anion of acid conc.]/[acid conc.] Make ICE table Ex.

HCNO

H+

CNO-

Initial

Molarity of remaining acid

0

Molarity of OH- in solution

Change

-x

x

x

Equilibrium

Molarity of remaining acid - x

0-x

Molarity of OH- in solution + x

Equation [x][molarity of OH- in solution+x]/[molarity of remaining acid-x]=10^-pKa 10^-pKa=Ka Solve for x after plugging in ka/pKa value and molarities. Ensure when doing logs you use Logbase10 (use symbolab + wolfram and compare) Plug previous x value into new formula to find pH Plug x value into: -log[H+]=-log[x] Use these instructions for base > acid Because there is more mols of base than acid remaining the mols of base remaining can now be calculated

A. Multiply molarity by volume of weak acid. Ex. .3300M * .1927L = A (mols of weak acid A) Make sure all volumes are in liters B. Multiply molarity by volume of strong base. Ex. 1.000M * .0266100L = B (mols of strong base B) Subtract A-B and see which remains. A-B=C C= remaining substance (in this case base) Take C (remaining base) and divide by total volume

C/Total volume in L = OH- in Mol/L Plug answer into new formula Kw=1.01*10^-14 (in Aleks) [H+][OH-]=Kw Solve for [H+] [H+]=Kw/OHPlug in Kw and your OH- you found and solve for H+ Ensure when doing logs you use Logbase10 (use symbolab + wolfram and compare) Take H+ answer and plug into pH formula pH=-log[H+]

Calculating the pH of a weak base titrated with a strong acid

A = Weak base B = Strong acid Check before doing problem Multiply molarity by volume of strong acid. Ex. .3300M * .1927L = A (mols of strong acid A) Make sure all volumes are in liters B. Multiply molarity by volume of weak base. Ex. 1.000M * .0266100L = B (mols of weak base B) Subtract A-B and see which remains. A-B=C C= remaining substance (in this case base) Because there is more mols of base than acid remaining the mols of base remaining can now be calculated Go down if acid > base Find molarity of unreacted base by: C/total volume in L = Molarity of remaining base Mols of cation produced = number needed for far right column of concentration in ice table Take previous B and do B/total volume in L B/total volume in L = molarity of thing in solution

Need to now solve for Kb Kb=[H+][cation of base conc.]/[base conc.] Make ICE table Ex.

Initial

C2H5NH2

OH-

C2H5NH3+

Molarity of remaining base

0

Molarity of thing in solution

Change

-x

x

x

Equilibrium

Molarity of remaining base - x

0-x

Molarity of thing in solution + x

Equation [x][molarity of thing in solution+x]/[molarity of remaining base-x]=10^-pKb 10^-pKb=Kb Solve for x after plugging in kb/pKb value and molarities. Ensure when doing logs you use Logbase10 (use symbolab + wolfram and compare) Plug previous x value into new formula H+=Kw/[OH-] See Kw value Solve for H+ Plug H+ value into equation to find pH: -log[H+]=-log[x] Use these instructions for acid > base Because there is more mols of acid than base remaining the mols of acid remaining can now be calculated

Multiply molarity by volume of strong acid. Ex. .3300M * .1927L = A (mols of strong acid A) Make sure all volumes are in liters B. Multiply molarity by volume of weak base. Ex. 1.000M * .0266100L = B (mols of weak base B) Subtract A-B and see which remains. A-B=C C= remaining substance (in this case acid) Take C (remaining acid) and divide by total volume C/Total volume in L = H+ in Mol/L Ensure when doing logs you use Logbase10 (use symbolab + wolfram and compare) Take H+ answer and plug into pH formula pH=-log[H+]

Calculating the pH at equivalence of a titration Multiply molarity of A by volume of A and get mols M*Volume in L of A = mols of A Divide mols of A by Molarity of B to give you remaining volume (needed to continue) A mols/Molarity of B = Other volume

Find total volume of solution Add together volume answer and volume of A given. Ensure all volumes are in liters Volume of A + Other Volume = Total V of solution Divide mols of A by total volume A/total volume in liters = Molarity of anion Make an ICE table Ex. 1 CNO-

HCNO

OH-

Initial

Molarity of anion

0

0

Change

-x

x

x

Equilibrium

Molarity of anion - x

x

x

Plug expression from table into Kb formula Equations are reversible (Ka Kb swappable) Kb=10^-pKb Kb=Kw/Ka [x^2]/[molarity of anion - x]=Kb if not given Kb explicitly must use bottom equation and substitute [x^2]/[molarity of anion - x]=Kw/Ka [x^2]/[molarity of anion - x]=Kw/10^-pKa Use this setup for instant answer x=rad((molarity of anion - x)(Kw/10^-pKa/b))

Use any and solve for x After solving for x plug values into new formula Only do next equation if PKa is given not PKb [H3O+]=Kw/[x] Plug in x and solve for H3O+ Plug H3O+ into -log[H3O+] to find pH -log[H3O+]=pH

Setting up a reaction table for a pH calculation with a common ion - ICE Tables Ex. 1 HF

F-

H3O+

Initial

.14

.35

0

Equilibrium

-x

x

x

Change

.14-x

.35+x

x

Setting up a reaction table is how you begin to find the equilibrium composition of a mixture of compounds among which a chemical reaction can take place.

Balancing a complex redox equation in acidic or basic solution Balancing complex redox reactions is often best done by the method of half reactions. Here's how it works. First identify what is being reduced or oxidized in reaction Ex. 1

Assign oxidation states for each element to easily visualize

Write half reactions of things being oxidized and reduced LEO GER Lose electrons Oxidation Gain electrons Reduction

Br2 went from 0 to -1 meaning it gained 1 electron = reduction MnO2 went from overall +2 to +5 meaning it lost 3 electrons = oxidation Because you know your electron changes you can now write half reactions with electrons included. Make sure to balance Bottom two reactions are for basic solutions.

Basic solutions will have OH- in half reaction with the more complex compound. Ensure balanced correctly After balancing half reactions multiply both reactions by a coefficient to make your electron values in each half reaction the same. Must cancel out

If in acidic solution see here

In acidic solution half reaction with more complex species will have extra H+ in it with H2O on right side. Make sure balanced After balancing half reactions multiply both reactions by a coefficient to make your electron values in each half reaction the same. Must cancel out Balancing Redox Reactions by Ion Electron Method

https://www.periodni.com/half-reaction_method.php

Analyzing a galvanic cell First identify what is being oxidized and reduced

See half reactions and do reduction and oxidation

Reduction always occurs at cathode Oxidation always occurs at anode To calculate cell voltage (emf)(cell potential) use following equation E=(reduction potential at cathode) - (reduction potential at anode) See reference given or aleks to get reduction potential values Use calculator to solve

https://www.periodni.com/half-reaction_method.php In Aleks take apart whole reaction and compare what you got to aleks data reference. Pick closest one and use.

Using the Nernst equation to calculate nonstandard cell voltage Nernst Equation E=(Esuperscript0)-(RT/nF)ln(Q) n= the number of electrons transferred from oxidants to reductants each time the redox

reaction runs F=Faraday constant R=gas constant R=8.31446 T=absolute temperature in Kelvins Reduction happens at cathode Oxidation happens at anode Start by taking apart equation and get half reactions Find reduction potentials of these half reactions Do E=(reduction potential at cathode) - (reduction potential at anode) to give you Esuperscript0 To find n value multiply the electron coefficients of your half reactions together. Only multiply if coefficients are are not divisible by one another.

To find Q you must do products over reactants molarities. Put reaction coefficients as exponents and solve for Q Make sure all temps are in kelvin Plug all found values into original equation and solve Units are J/C in aleks

Calculating standard reaction free energy from standard reduction potentials deltaG=-nFE n=number of electrons that are transferred during the reaction. F=Faraday constant E=emf

Find your half reactions of redox and see reduction potentials Use formula to find Esuperscript0/emf E=(reduction potential at red.) - (reduction potential at ox.) To find n value multiply the electron coefficients of your half reactions together. Only multiply if coefficients are are not divisible by one another. After plugging into original equation answer will come out as C*V Convert C*V to KJ by taking your scientific notif answer and move decimal over 2 places

Using a rate law Rate law is a law that tells you how the rate of a chemical reaction depends on concentrations of its reactants the sum of all the reaction orders is called the overall reaction order.

A=Molarity of A B= Molarity of B May need to solve for^ when an exponent is 1 we don't write it. That means that when a rate law has the concentration of a reactant with no power, the reaction is first order in that reactant. if the molarity is halved the initial rate must be multiplied by one-fourth since half raised to power of 2 is ¼ . When altering molarity by a factor you are to put that factor to the original power. If molarity is multiplied by a number the rate will be multiplied by that number raised to the power To solve for K just rearrange equation rate/[A][B]=K If asked what would happen to a value if multiplied by a factor Raise new factor to your exponent of the substance and then multiply your initial value by this number Ex.

Calculating the pH of a buffer Use either of the following to determine pH of a buffer

Ka=10^-pKa Kb=10^-pKb Kw=10^-pKw Kw=1*10^-14 pKw=14 at 25 celsius/standard conditions HA=concentration of weak acid A-=concentration of conjugate base BH+=concentration of conjugate acid B=concentration of weak base Plug in values and solve for pH Calculating the composition of a buffer of a given pH

To solve you will use henderson hasslebach equations

Rearrange equations so that you can calculate desired equilibrium molarity of conjugate acid or whichever molarity value is missing.

Ex. After solving for missing concentration you will multiply this value by the desired liters of buffer to give you the mols of solvent needed to make the buffer

Multiply mols by molar mass of thing and get grams of thing needed Deriving Kb from Ka Ka*Kb=Kw Kw=1.01*10^-14 (in aleks) at 25 celsius Acids gain an H+ and bases lose an H+ Calculating the pH of a salt solution - Bronsted lowry acid part First identify if your substances act as bronsted lowry bases or acids Weak bases = bronsted lowry acid Weak acids = bronsted lowry base If acids then they will release H3O+ If bases then they will release OHFirst need to make sure you have Ka value or Kb value needed for equation - If given a weak base need to find Ka. If given a weak acid need to find Kb Ka=Kw/Kb Kb=10^-pKb Ka=10^-pKa Once you have Ka/Kb value make an ice table with the concentrations given Ex.

Since only one conc. Is given the others will be 0 Form an equation from this. Products over reactants = Ka/Kb value. Solve for x. Use this X to plug into pH equation and solve

pH=-log[H3O+] If working with bronsted lowry base then see bottom directions

Follow same steps as before but before plugging your value into log equation you must solve for H3O+ Bases release OH- and must be turned to H3O+ H3O+=Kw/OHAfter solving for H3O+ plug into pH equation pH=-log[H3O+] Calculating solubility Solubility of a solvent is equal to its concentration in a saturated solution. First find the mass of solvent that has dissolved in the solution. Subtract initial mass from remaining mass. Take this mass and divide by volume of solution Using Henry's Law to calculate the solubility of a gas Henrys equation C=KhP Kh= henrys constant P= partial pressure C= solubility of gas To find mass of thing that will dissolve do the following Multiply henrys constant by partial pressure to find solubility of gas C=KhP Take solubility of gas and multiply it by the volume given to you C*(V in L) = mols of thing Multiply mols by molar mass and you have your answer

Calculating standard reaction free energy from standard free energies of formation To solve these you will need the standard molar free energies of formation. In Aleks can be found in data table. standard molar free energies of formation = deltaGf Elements in their standard state are considered to have a energy of 0 To find standard reaction free energy of the reaction you must add up the total amount of gibbs free energy of the products and subtract the total amount of gibbs free energy on the reactants side deltaGf=(gibbs of product) - (gibbs of reactants) Make sure to include stoichiometric coefficients when calc. Gibbs free energy

Calculating the pH of a strong acid solution Strong acids dissociate completely 100% - important detail Knowing that it is strong acid allows us to be able to calculate final H3O+ molarity First you will need to calculate initial mols of acid. Divide mass of acid given to you by molar mass of acid to find mols. Make sure units are in grams

Next you must calculate mols of H3O+ produced. Pay attention to original reaction equation - ensure your stoichiometric ratio is 1:1. If not change it and apply. (x produced/y consumed) Multiply your mols of acid by your ratio to find mols of H3O+ produced

To calculate final molarity divide mols of H3O+ produced by volume in liters of solution

Finally to get pH you will take your mol/L calculation and plug into Ph formula pH=-log[H3O+] Calculating the pH of a strong base solution First you will need to calculate initial mols of base. Divide mass of base given to you by molar mass of base to find mols. Make sure units are in grams

Next you must calculate mols of OH- produced. Pay attention to original reaction equation - ensure your stoichiometric ratio is 1:1. If not change it and apply. (x produced/y consumed) Multiply your mols of base by your ratio to find mols of OH- produced

To calculate final molarity divide mols of OH- produced by volume in liters of solution

Once you have mols of OH- produced you must now solve for mols of H3O+ H3O+=Kw/OH-conc. Divide Kw by OH- and solve for H3O+

Finally to get pH you will take your H3O+ calculation and plug into Ph formula pH=-log[H3O+] Using an integrated rate law for a first-order reaction Using the integrated rate law you can calculate the change in concentration of a reactant over time. Below are the rate laws for a first order reaction. Other orders are different.

A=concentration of A [Ao]=initial concentration of A K=rate constant t=time given Plug in current conc. And initial conc. And solve Calculating the Ka of a weak acid from pH The key to solving this problem is to use the acid dissociation constant formula. Ex.

This equation can be used to calculate Ka from the molarity concentrations already given to you. Form an ICE table with the values given

Since we are doing an acid we will dissociate H3O+ meaning x will be equal to the H3O+ dissociated. (for bases must do Kw/OH- to find H3O+) To solve for x in this case we will do the following equation x=H3O+=10^-pH This means x is equal to 10^-pH After solving for x plug it back into ICE table and form the Ka constant equation. With all concentrations in you can solve for Ka

Using a second-order integrated rate law to find concentration cha...