chem ch.16 questions from book

Remember that [H] x [OH] = kw

kw = 1.14e-14

So if we only have kw here , and we see [H]+ as X, and (OH) as X

That’s the same as

X² = 1.14e-14

we can solve for x

In this equation, you’ll set up an ice table and find that it ends with

Ka = (2.5e-5)² / X-0.1

Because the value for [H]+ was so small, JUST LIKE WE DO FOR Ka,

• You can remove the X to find Ka.

x

If both the components are a strong base and strong acid component the pH will be…

7

1. Count the O’s, more O’s attatched = higher strength

2. See other atoms

F > Cl > Br> I

O > N > C> H

For strength!

Important Count oxygen groups first, this is the biggest dealbreaker for strength.

a

Remember these

1. As the strength of the bond to H increases, the strength of the acid → decreases

• Strong H–X bond → harder to lose H⁺ → weaker acid

2. As the polarity of the bond to H increases, the strength of the acid → increases

• More polar bond → H⁺ more easily released → stronger acid

3. As the size of the anion bonded to H increases, the strength of the acid → increases

• Larger anion → negative charge better stabilized → H⁺ easier to lose → stronger acid

a

When solving a polyprotic equation, what’s the point of using K1 vs K2?

1. You write out the full expression of lets say

H3C6O2 acid

for k1

H3C6O2 + H2O —> H2C6O2 + H3O

For k2

H2C6O2 + H2O —> HC6O2 + H3O

For k3

HC6O2 + H2O —> C6O2 + H3O

So if you’re asked to find the concentration of C6O2, this is where we’d have to use K3 to find it, we’d need to make 3 seperate ice tables to find the concentration of C6O2.

• For pH, you can usually just use the value of H3O+ at k1.

stopped at 16.9, finish 16.9 and all of 17.0 except the portions she said not to tomorrow. Then just review.

What are 2 traits to look for in a Lewis base?

negative charge

lone pair

What traits to look for in a lewis acid?

positive charge

incomplete octet -

Check for a periodic table if we’re gonna get one on the exam, because the image explains it pretty clearly based on valence electrons, but you need a periodic table to know that.

memorise these i guess

d

Make ice table, it’s unique for this one because when k is small for this example

we can do ka (given just didnt write it) = (20 + x) (x) / (30 - x)

and it becomes

Ka = (20)(x) / 30

so we eliminate the x that’s combined with a value in both the numerator and denominator.

a

A buffer forms on weak acids and bases between a conjugate acid (1 more H+) and its conjugate base (has 1 less H+)

only B is forming a buffer because there’s a 1H+ difference between them

Note, any time there’s a strong acid or base present, a buffer can’t form.

So HI and I, even though they’re conjugate base + acid, they can’t form a buffer because HI is strong

x

Adding products shifts to left

(remember adding reactants right, decrease reactant left, increase products left, decrease products right

So PCl5 goes up, PCl3 goes down, Cl goes down

Ammonia =

Ammonium ion =

Ammonia = NH3

Ammonium = NH4

Should be NH3 + h2O —> NH4 + OH

list the steps for solving and the setup

1. % ionization (0.01, which is 1% as decimal) = H3O+ eq / acid initial (0.1). don’t include 100.

2. we solve for H3O, then go back to the balanced equation to see the concentrations for HZ and Z so we can plug into Ka

Find the concentration of OH ions in each of the following, sometimes it may be necessary to use Kw to solve.

A. Find H3O with pH —> Use Kw to solve for OH once you have H+

B. Strong acid disassociates completely, so H+ = HNO3 conc, then use Kw to solve for OH

C. Convert pOH to OH-

D. Water : 1e-14 = [H][OH) is the same as 1e-14 = X² , so we solve for x to get the conc of OH

In the HH equation

What’s unique about the values that can be used for log(base/acid)

moles can be used instead of just M

x

1. Find pKA

2. Plug given pH in

3. simplify

4. do 10^(-)value to get the ratio of ion to acid.

For a. just calculate the pH by doing

pka = -log(ka)

Then do log (base/acid)

For b. write out a reaction

Weak acid + strong base (OH) —> conjugate acid + h2o

• make BAA chart

solve

M1V1 = M2V2

(0.05L) (0.5M KOH) = (0.5L) (M2)

Solve for M2, this is the new concentration of OH after adding, because KOH disassociates completely.

Then use that to solve for pH, because you have conc of OH- ions.

d

Liam for titration

If starting with weak acid + strong base

do weak acid + OH —> Weak base conjugate(minus 1H) + H2O

if starting with weak base + strong acid

do weak base + H3O —> weak acid conjugate (+1H) + H2O

If it’s Strong acid + Strong base

Do H+ + OH- —> H2O

General rule of thumb, never include the full equation for a strong base or acid, just simplify to H+ or OH

1. Convert initial molality of Acid and Base in buffer to moles, i.e do 0.250M of acid x 0.500L to get moles

2. Calculate moles of new substance’s OH or H+ by doing the ml added x the M it contained

3. Write out in ice table, subtracting the new substance value found in step 2 from the new initial substance found in step 1

4. Do the new EQ concentration / Total volume (volume used in step 1 + volume in step 2)

5. plug into the henderson pH equation.

What does the X and Y axis represent?

What do regions A, B, C, and D represent?

Strong Acid-Strong base titration

X axis = volume of base added in ml

Y axis = pH

A = initial pH of acid alone. it can be found by doing -log[H3O]

B = after base has been added but before the EQ point. This pH can be calculated from the conc of H+ that hasn’t been neutralized

C = EQ point at pH = 7.0

D = after EQ point where excess base determines the pH. First calc the pOH from the conc of excess OH-, then calc the pH.

SA - SB titration

1. calc pH by doing -log(0.300)

2. For b.

First, convert the molality to moles of HCl, and moles of NaOH

this is done by doing Molality x Liters of solution

After you do this, plug them into Ice Table, Leave the initial value of the acid and the inital value of the base, and subtract do Initial H+ - Initial OH, this will give you an EQ value in moles. Divide that EQ value by the total amount of volume in the solution, to get it into molality. From here, you can calculate pH

c. pH of a titration of a Strong Acid and Strong base will always be 7

D. First, multiply the moles of the base by the volume you’re now adding

• Then, the value you just obtained in part D, you are going to subtract the initial concentration found in part A from it.

• After this, you will divide by the total liters added of the base

• This will give you an answer in OH-, —> convert to pOH —> convert to pH

The following shows

Titration of a strong base with a an acid

The following shows

Titration of a strong Acid with a strong base

SA-SB
Strong acid, with stong base added

SB-SA
Strong base, with strong acid added.

What’s the following show?

WA-SB

a. Inital pH
B: Buffer range where some conjugate base has been produced but not all WA has been consumed

C: Equilavence point where pH is more than 7

D after equilavence point where excess base determines pH

Determine which will have highest pH at EQ point of titration given pKA between several acids

The higher the pKA, the weaker the acid, and the higher the pH.

a

phenol, highest pKA, highest pH, weakest acid.

ammonium ion, pkA, middle pH,

hypobromous acid, lowest pKA, strongest acid.

a

Step 1: Before NaOH added pH

1. Write out HA ←→ H + A

• Make an Ice table to calculate pH using the given Ka

2. Half of the equilavence point

Rewrite equation as

HA + OH —> A + H2O

Show me the exact math yo

value (i.e -2) = log (base/acid)

How do you solve for the base and acid?

10^-2 gives you ratio of base to acid, as it will cancel the log

-2 = log(base/0.750)

How do you solve for moles of base?

10^-2 will cancel the base, and become 0.01

0.01 = X/.750

solve for X

A. 0.0500mol NaOH in 1L of water

1. Find [OH-] M

0.0500mol NaOH/ 1L = 0.0500 M

From here, you can find pOH —> PH

For solving this problem, what’s the trick that we want to look for?

Find a value where pKA = pH

Buffer works best when pH = pKa

What do you do to solve this?

Find a pKA that matches 7.4

Step 1, titration

What’s the first step for all titrations

Step 2, calculating pH before adding any acid or base, what do you write out?

Step 3, calculating pH after adding base. What is it for

SA + SB

vs WA + SB?

Step 4: Calculating pH at EQ for SA + SB

Step 5 Calculating pH at halfway point for WA + SB

Step 6: Calculating pH at Eq for WA + SB

step 1: M_aV_a = M_bV_b , this gives you the volume at the Equilavence point.

Step 2: Calculate pH using an Ice table, simply write the disassociation of the SA or WA and calculate with an ice table (if it’s a WA it will form a conjugate base + H+). If it’s a SA, it will match the [M] of the H+.

Step 3. Any pH before EQ point

• Rewrite expression for SA + SB as

OH + H —> H2O

B

A

A

Rewrite expression for WA + SB as

WA + OH —> H2O + Conjugate base of WA

B

A

A

4. pH at EQ for SA + SB = 7

5. pH at halfway point for WA + SB = pKa

6. pH at Eq point for WA + SB

WA + OH —> H2O + Conjugate base of WA

• you’ll eliminate all OH + WA and be left over with conjugate base of WA

• Take the moles of Conjugate base of WA produced, and divide by total volume of WA + OH added to get it in (M)

Create an Ice table showing how the Conjugate base reacts with water

Conjugate Base WA + H2O —> OH + Conjugate Acid of WA

• Use Kb instead of Ka, as you’re solving for a base

• Solve for X

• Convert X to OH, and solve for pH

Any amount past the EQ point for SA-SB or WA-WB

• take the leftover from the

B

A

A

and convert it into (M) then into pH

A buffer consists always of a

conjugate acid and its base