Answers to Frequently Asked Questions

Table of Contents

  1. Reviews from 101
  2. Solutions
  3. Thermo including absolute zero
  4. Kinetics
  5. Equilibrium problems
  6. what is the 100 times rule (equilibrium... ?
  7. What solubility rules do we need to know ... ? (includes list of polyatomic ions to memorize)
  8. pH of a salt
  9. Henderson?
  10. What reaction do you use when adding a strong acid or strong base to a buffer?
  11. Titration problems
  12. pH in an acid/base titration
  13. How do I balance redox equations?
  14. Electrochem (including products of an aqueous salt, Faraday's law, overvoltage, batteries)
  15. Nuclear Particles

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Reviews from 101...

Can you repeat the naming rules and give us the list of 10 ions to memorize that you gave your chem 101 class?

Click here for the naming rule and the list of 10 ions to memorize.


What is an equivalent?
In acid/base rxns, an equivalent supplies or reacts with 1 mol H+ or OH- and in Redox loses or gains 1 mole e-. Therefore, in H2SO4 you have 2 equivalents per mole of the compound, as you do in Ba(OH)2. These can supply the ions. For salts, you can use the reacts with (or came from which acid or base model): For example: Na2SO4 have 2 equivalents per mole of the compound, while Ca3(PO4)2 has 6 equivalents per mole.

How are Normality and Molarity related?
Molarity is: M = moles / liter soln, while Normality is: N = equivalents/ liter soln. To convert between the two, you need the number of equivalents per mole of the solute.
Therefore: 6N H2SO4 is 3M H2SO4. You could calculate this by: 6N is (6 equiv/ Liter) x (1 mole H2SO4/2 equiv) = 3 mol/L , so 3M
And 3.5 M Na2SO4 is 7M Na2SO4 using this calculation: (3.5 mol/ Liter) x (2 equiv/ 1 mole Na2SO4) = 7 equiv/ L , so 7N

You Try: 2M NaCl = ___N NaCl; 6M H3PO4 = ___N H3PO4; 4N NaClO3 = __ M NaClO3; 8N Co(OH)2 = __ M Co(OH)2

answers: 2N, 18N, 4M, 4M

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What is a state function?
Answers to this problem depends on the context. A state function is a variable that defines the state of a system; it is a function that is independent of the pathway by which a process occurs. This implies a process, and that a state function only depends on the initial and final values.


Why should we use 273.15 for the temperature in Kelvins instead of 273?
273.15 is more accurate (5 sig figs) than 273 (3 sig figs). The decision to use the less accurate 273, depends on what other numbers in the problem are limiting the sig figs.

What is the difference in delta H , delta Ho, and delta Hof

delta H is for the reaction as it is written. This can be at the temperature specified and for any pressure or molarity.

delta Ho is the delta H at standard conditions(standard enthalphy change). This means at 25 degrees C, with all aqueous solutions at 1 M and all gasses at 1 atm. Also any elements will be at their standard phase (solid, liquid, or gas).

Even more restrictive is
delta Hof (standard enthalphy of formation). This is for an equation in which elements in standard conditions (temp, phase of matter) make 1 mole of a compound.

Can we get a copy of the overhead about Temperature and delta G?
click here for a pdf

Can there be a negative value for absolute temperatures (temps below absolute zero)?

Read down to the part about how it will affect entropy in this link!
For a video on the concept:, see this link!

However, consider these comments by a physical chemist:
It has been well known for a long time that a system in a population inversion could be argued to have a negative absolute temperature. So the active medium in (almost) any laser could be said to have a negative temperature. And there is no doubt that such a negative temperature is in a higher energy state than any positive temperature state of the same system. But it all depends on the meaning of the word temperature. Most conventional definitions involve a system at equilibrium, which largely precludes the inverted population. this may not be a real advancement or feat.

**Click here for the joke: Using thermodynamics, how hot is hell?

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**The reading on integrated rate laws from another textbook is on the protected page. (The page where the ppts and reviews are found).

** Click here for a copy of the integrated rate problems. Click here for a complete solution.

**Want another problem set for integrated solutions?  Click here.  Solutions are at the end.

** Want a question on Mechanisms? Click here for a pdf

1. catalyst is response (C) or P;
2. intermediate is response (D), note all responses for #1 are also intermediates except the catalyst P; responses for #2 are reactants or products, except the (D) response.
3. rate law is response (A) or rate = k [ABQ] [P]

**A half-life is the time for 1/2 the conc to be used up. If the rxn is first order, then each half-life will be the same amount of time. If the rxn is zero order, the time for the second 1/2 life will be LESS than that of the first 1/2 life. For second order, then second half life will be MORE time than the first 1/2 life.

**The integrated rate equations call for molarity, but you can use mass (like mg) for radioactive decay, which is always first order.

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 Equilibrium problems

Do you have more equilibrium problems? Click here for a practice sheet on equilibrium (answers at the end). FOR SPRING 2018 EXAM 2, OMIT QUESTIONS 3-5, AS THESE DEAL WITH LE CHATELIERS LAW. But do 3-5 for EXAM 3.

Try the equilibrium simulator at: or at

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What is the 100 times rule (equilibrium)..?

What is the 100 Times Rule?

When you are given initial conc and the K, you make a I,C,E table using X values for the unknown amount that reacts. Suppose you ended with the following for K in a reaction of
A <--> B with the initial of 0.33 M of A, and K = 2.0 E-11.

K = 2.0 x 10 e-11 = X/.33 - x

The 100 times rule says to use 100K = x in the places where x is subtracted or added to an original conc. If the conc would be the same as the original when considering sig figs, drop the - x part. In this case, 0.33 - (2.0 E-9) would still be 0.33 M when using 2 sig figs, so you can estimate:

2.0 x 10 e-11 = X/.33

now solve for x. You can sub the value in for x into the equilibrium conc from your table. You were correct to drop the -x if 0.33-x is 0.33 to 2 sig figs.

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What solubility rules do we need to know... ?

I was just wondering which solubility guidelines we need to memorize in order to determine if a salt is soluble or not.

Good Question! I usually ask students to know the basics (almost all sodium, potassium, ammonium, chloride, and nitrate salts are soluble.) If you need more than that, the problem will say "soluble potassium hydroxide" or KOH (aq) or will give you a solubility table.

**Click here for a handout on naming rules and the ions to memorize from chem 101

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How do you determine the pH of a salt?

For the examples: First split the salt into the cation and anion.  Then find the parent of each ion and label the parent as SA, SB, WA (weak acid), WB (weak base), or N (neutral) . Next decide if the ions are WA, WB, or N.  Finally, decide the pH of the salt from combining the 2 ions.

Cation from base
(parent base)
Anion from acid
(parent acid)
pH of Salt Salt
NH4+ = WA WB (NH3) Br- = N SA (HBr) Acidic salt NH4Br
Na+ = N SB (NaOH) NO3- = N SA (HNO3) Neutral salt NaNO3
Na+ = N SB (NaOH) CHO2- = WB WA (HCHO2) Basic salt NaCHO2
NH4+ = WA WB (NH4OH) HCO3- = WB WA (H2CO3) Depends on K's NH4HCO3
C5H5NH+ = WA WB (C5H5N) NO3- = N SA (HNO3) Acidic salt C5H5NHNO3

In Summary for pH of Salts:

Cation from base Anion  from acid Salt
N SB N SA Neutral salt
N SB WB WA Basic salt
WA  WB N SA Acidic salt
WA  WB WB WA Depends


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Henderson Equation?

How do you know when you can use the Henderson-Hasselbalch equation and when not to?

The Henderson equation can be used:

If the acid is monoprotic
if the salt is univalent (the anion & cation -1 and +1 respectively)
If both the acid and salt are in "reasonable concentrations" according to your book. Some books define 'reasonable" as equal to or larger than 0.05M, others list less than 1 M.

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What reaction do you use when adding a strong acid or strong base to a buffer?


Weak Acid/conjug. buffer Weak Base/conjug. buffer
When SA is added:

Reaction is:

SA + salt --> WA

H+ + X- --> HX

HCl + NaAc --> HAc + NaCl

H+ + Ac- --> HAc

When SA is added:

Reaction is:

SA + WB --> salt + water

HCl + NH3 --> NH4Cl

H+ + NH3--> NH4+

When SB is added:

Reaction is:

SB + WA --> salt + water

NaOH+ HAc --> NaAc + H20

OH- + HAc --> Ac- + H20

When SB is added:

Reaction is:

SB + salt --> WB + water

NaOH + NH4Cl --> NH3+ H20+ NaCl

OH- + NH4+ --> NH3+ H20

Can you give us a sample problem for the addition of a strong Acid to a buffer?

To 150 mL of a buffer soln that is 0.334 M ammonium bromide and 0.342 ammonia, add 0.0369 mol HBr. Give the pH. Click here for a pdf solution.

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Titration Sample Problems

Find the pH of 100.0 mL of 1.0000 M acetic acid titrated with 0.150 M NaOH to the equivalence pt.

Find pH if 200. mL of 0.25 M HCN is titrated with 55.0 mL of 0.25 NaOH.

First react the HCN + NaOH --------> NaCN + H2O in moles
            (.200L)(.25M)  (.0550L)(.25M)
        I        0.050 mol      0.01375 mol
change       -.01375        -.01375        .01375               
Final            .03625            0                .01375mol

So you end with a weak acid and its salt  
Change to molarity and use the Henderson:
.03625mol HCN/.255 L = 0.142 M HCN    0.01375mol/.255 L = .0539 M NaCN


Henderson:  pH = pKa + log [(salt)/(acid)]
pH = -log 4.9 e-10 + log (0.0539/0.142)
pH = 9.31  +  (-.4207) = 8.89  in 2 sig figs

Problem set for acids, bases, buffers, titrations?  click here  (last 2 questions are just a math review, others are on topic)
Answers to the problem set?  click here

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How do you find the pH in an acid/base titration?

pH of acids titrated BY BASES

  SA + SB WA + SB SA + WB
Initial pH [SA]--> [H+] [WA] <--> [H+] equil. [SA]--> [H+]
pH before e.p. 2 step:
react A/B,
[excess SA] --> [H+]
2 step:
react A/B,
equil with excess [WA] + [salt]
2 step:
react A/B,
[excess SA] --> [H+]
pH @ e.p. pH = 7 2 step: react A/B, equil with
[B salt anion]<-> [OH-]
2 step: react A/B, equil
[A salt cation] <--> [H+]
pH final 2 step: react A/B, excess [SB] = [OH-] 2 step: react A/B, excess [SB] = [OH-] 2 step: react A/B, equil with [WB] + [salt]

pH of bases titrated BY ACIDS

  SB + SA WB + SA SB + WA
Initial pH [SB]--> [OH-] [WB] <--> [OH-] equil. [SB]--> [OH-]
pH before e.p. 2 step:
react A/B,
[excess SB] --> [OH-]
2 step:
react A/B,
equil with excess [WB] + [salt]
2 step:
react A/B,
[excess SB] --> [OH-]
pH @ e.p. pH = 7 2 step:
react A/B,
[A salt cation] <--> [H+]
2 step:
react A/B,
[B salt anion] <--> [H+]
pH final 2 step:
react A/B,
pH -> excess [SA]
2 step:
react A/B,
pH -> excess [SA]
2 step:
react A/B,
equil [WA]
+[B salt]
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How do I balance redox equations?

Balancing Redox Equations: 1/2 reaction method

1. Use unbalanced reaction to write two half-reactions.

2. Balance elements except O and H

3. Balance O by adding H2O

4. Balance H by adding H+

5. Balance charges by adding electrons

6. Multiply each 1/2 rxn by coefficients to get the same # of electrons in each

7. Add the two 1/2 rxns

8. If in basic solution: Add OH- to each side of the equation to transform all the H+ to H2O, then cancel H2O that appears on both sides.

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What are the products of the electroylsis of aqueous CuBr2?

The problem is that these are aqueous. So you must consider all possible oxidations and all possible reductions and choose the ones with the more positive voltage:

2Br - --> Br2 + 2e- Voltage= -1.06 ****** most positive oxidation (less energy required)
2 H2O --> 4H+ + O2 + 4e- Voltage= -1.23

Cu2+ + 2e- --> Cu Voltage= -0.34 ****** most positive reduction (less energy required)
2 H2O + 2e- --> H2 + 2OH- Voltage=-0.83
Answer: Cu + Br2

I was also wondering why the reduction and oxidation equations of water are completely different? For every other equation, we just flipped the reduction equation from the Standard Reduction Potential Table and then obtained the oxidation reaction. Why is it that there are two separate equations for water and different voltages?

The problem is that with an aqueous salt, you only have the ions from the salt and water in the container. These are the only reactants you can have. If you just used one equation for water for both the oxidation and reduction, then you would not have water as a reactant in both. This is the reason that 2 equations must be used.

I don't really understand the concept of over voltage~ could you explain again with some examples please?

For gases such as oxygen and hydrogen to be produced at metal electrodes, there seems to be additional voltage required. This extra voltage is called overvoltage and results from kinetic factors. The overvoltage is about 0.4 - 0.6 volts. So the reduction of water:
2H2O + 2e- ---> H
2 + OH- has E = -0.83 v is really -1.43 volts with metal electrodes . This explains why Cl2 gas is produced with the electrolysis of aquesous NaCl. Cl- (-1.36) is oxidized instead of water (-1.23). If the conc of Cl- is high and if you consider the overvoltage of water (making it -1.83), then Cl- will be oxidized to Cl2.

I will not hold you responsible for using overvoltages, only for knowing that they exist. They were brought up to explain why the aqueous NaCl example in the text did not give the expected products.

When to switch???  when both reduction voltages are positive, do we switch the least negative? This seems to be one of the most confusing parts of the whole chapter. Please help clarify my confusion.

The best rule to use, is if the reaction is spontaneous or a voltaic cell, the most negative voltage switches to an oxidation. So if both are positive ( say 0.80 and 0.34) switch the most negative or the on closer to negative on the number line (so switch the 0.34 to -0.34). If both are negative (say -0.66 and -0.47), then switch the most negative (switch -0.66 to 0.66).

Faraday's Law?? Using a 55.0 amp current for 5.0 hours in molten AlCl3. Calculate the mass of Al metal produced.

55.0 amp 5.0 hours 60 minutes 60 sec 1 mol e- 1 mol Al 27.0 g = 92.3316 g or 92.3 g in 3 sig figs
    1 hour 1 minute 96,500 amp*sec 3 mol e- 1 mol al-  

The idea that a Faraday is 1 mol of e- was important.

Batteries of the future:

Hear this talk:

How do you answer the tutored problem 21.2.1 on assignment 21.2a parts c and d:

(c) Will I2(s) oxidize Zn(s) to Zn2+(aq)?

What this you wants you to do is compare the voltages to see if you get a positive voltage (meaning spontaneous).
The problem is wanting I2 and Zn(s) to be the reactants. You also need to ck that you have a reduction and an oxidation with these as the reactants.

You are given this:
Half-reaction------E° (V)
I2(s) + 2e- --> 2I-(aq) with 0.535V
Sn2+(aq) + 2e---> Sn (s) with -0.140V
Zn2+(aq) + 2e- --> Zn (s) with -0.763V

For I2 to be the reactant, then your reduction is I2 + 2e- —> 2I- with a voltage of +0.535V
For Zn to be the reactant, the oxidation is Zn —> Zn2+ + 2e- with a voltage of +0.763V

These two represent an oxidation and a reduction, these would make an overall reaction that would have a voltage of 0.535 + 0.763 positive overall voltage means that this reaction would react spontaneously

(d) Identify all species that can be reduced by Sn(s). If none, leave box blank.

What this you wants you to do is to see if you can have Sn(s) as the reactant, which would make the oxidation as: Sn -> Sn2+ + 2e- with a voltage of 0.140

The question you are trying to answer is which reduction would give you a positive voltage when you added the reduction to the oxidation.

IF Zn2+(aq) + 2e- --> Zn(s) with -0.763V was the reduction, then you would end up with a negative voltage,
so non spontaneous. (0.140 + -0.763)

IF I2(s) + 2e- --> 2I- (aq) with 0.535V was the reduction, then you would end up with a positive voltage,
so spontaneous (0.140 + 0.535). Therefore I2 can be reduced by Sn.

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Nuclear Particles mentioned postron beta and electron beta, I know that the signs are different (+ and -) but how do I know when to use what when it just says beta?

A beta particle is a high energy electron. A positron has relatively no mass & a positive 1 charge (think of it as a positive electron).

**A half-life is the time for 1/2 the conc to be used up. If the rxn is first order, then each half-life will be the same amount of time. If the rxn is zero order, the time for the second 1/2 life will be LESS than that of the first 1/2 life. For second order, then second half life will be MORE time than the first 1/2 life.

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