Acid Base equalibria

Question 1

What is a bronsted-Lowry ACID

What is a bronsted-Lowry BASE

Answer 1

Acid is a proton donor (donates H+)

Base is a proton acceptor (accepts H+)

Question 2

What is a lewis acid?

What is a Lewis base?

Answer 2

Lewis acid is an electron acceptor

Lewis base is an electron donor.

(notice that B is sp2, meaning there is an emplty unhybridized p orbital)

Question 3

1) Why is water amphoteric?
2) What is the concenration of hydronium ions in water?
3) What is the concentration of hydroxide ion in water?
4) What is the unit of concentration

Answer 3

1) It can act as an acid or base.

2-3) Both hydronium and Hydroxide concentration are 1x10^-7. Because the concetration of both are the same the solution is neutral.

4) Concentrations are in Molarity

Question 4

What is a solution with more Hydronium ions than Hydroxide ions called?

Answer 4

Acidic. pH less than 7

Question 5

What is a solution with more Hydroxide ions than Hydronium ions called?

Answer 5

Basic, pH greater than 7

Question 6

Find the concentratin of hydroxide ions in a solution of lemon juice that has a hydronium concentration of 2.2x10^-3

Answer 6

(2.2x10^-3)(x) = 1.0 x 10^-14 => 4.5x10^-12

Question 7

Caclulate the following pH:

Hydronium concentration is 1.5x10^-4

Answer 7

take the - log. 3.82. Or you can see that the power is-4 which means if the log as 1.10^-4 would be 4, it should be close.

Question 8

Calculate the concetration of hydronium ions if the pH is 3.82.

Answer 8

take 10^-3.82 = approx 1x10^-4. The exact answer is 1.5x10^-4

Question 9

Calculate the pH of an aqueous ammonia solutions with a hydroxide concentration of 2.1 x 10^-3

Answer 9

You can use either equation:

pOH + pH = 14 or (Concetration of OH)(Concetration of H30)=1.0x10^-14

without a calculator:

pOH of 2.1x10^-3 is approx 3. Therefor, 3+x=14, pH=11

With a calulator:

pOH of 2.1x10^-3 = 2.68, therfore, 2.68+x=14, pH= 11.32

Question 10

Caclulate the pH of a 0.030 M HNO₃ solution

HNO₃ + H₂0 ->

Answer 10

HNO₃ is a stong acid, all molecules will react..

HNO₃ -> H+ NO₃-

Because the concentration of HNO₃ completely reacts, the concentration of H is a 1 to 1. therefore, the conc. is 0.030M

pH= -log(H)

without a calculator:

pH= -log(3x10^-2) = 2

with a calculator:

1.52

Question 11

Caclulate the pH of a 0.20 M solution of NaOH.

Answer 11

NaOH is a strong base, so it will 100% dissasociatio into

Na+ and OH-

2 ways to do this:

without a calculator:

-log(2x10^-1) is approx 1 for pOH. 1+pH=14, pH=13

With a calc.

-log(0.20) = 0.7,=pOH. .7+pH=14,pH= 13.3

Question 12

calculate the pH of an aqueous solution that contains .11g of Ca(OH)₂ in a total volume of 250 mL.

Answer 12

pH = 12.05

1) Write the chemical equation:

Ca(OH)₂ -> Ca²⁺ 2OH^-

Notice, there is double the hydoroxide ion for each Ca(OH)₂.

2) Calculate Molarity/ concentration:

add up the atomic weight from the periodic table, Ca=40, O =16(x2), and H =2, (x2), therefore there are 76 g/mol

1 mol x .11 g = .0014 mol

76g

the .0014 molar mass is for the reactant, but since there are double the OH, the M is actuall .0014*2= .0028 mol

3) Caclulate Molarity, (Mol/ L) 0.0028/0.250L = .0112
4) Use any of the standard equations to calculate use this info in the standard pH/ pOH equations
- log( 1.1^-2) = approx 2 (1.95 with a calculator), = pOH,now calculate pH. pH+pOH=14, pH+1.95=14, pH = 12.05

Question 13

What is Ka?

What is pKa?

Answer 13

1) Ka is the ionization concentration. The larger the Ka value, the more dissociation of the molecules in solution and thus the stronger the acid.

Ka=(H⁺)(A^-)

 (HA)

2) pKA is the -log(Ka)

The smaller the pKa the stronger the acid.

Question 14

Calculate the pH of a 1.00 M solution of Acetdic acid, CH₃COOH _(aq)

Ka= 1.8x10^-5

Answer 14

1) write the equation

CH₃COOH + H₂O H₃O + CH₃COO^-

2) Figure out the proportions:

Before reaction takes place

1.00M 0 + 0

After reaction

1.00M -X X +X

Ka= ( H₃O )(CH₃COO- ) (CONCENTRATION OF EACH)

         CH<sub>3</sub>COOH

Ka= _(x)(x) _ = 1.8x10^-5

   1. 00-x 3) could solve for x using the quaradic equation

OR

you could say that 1.00-x when x is a really, really small number = 1.00, therefore equation turns into Ka=x²/1.00

1. 8x10^-5 = x² = approx .004 (concenration of H₃O⁺)
   4) take the - log of .004 = 2.38

Question 15

1) what is Kb
2) How to cacluate Kb?
3) What is pKb

Answer 15

1) Base ionization constant. base disassocation constant.

Base + H²O BH^{+ ;}+ OH^-

2) Kb = (BH⁺)(OH^-)

B

3) - log (Ka)

Question 16

Calculate the pH of 0.500 M solution of NH₃

Kb of amonium is 1.8x 10^-5

Answer 16

1) Write the equation:

NH₃+ H₂O NH₄+ OH^-

.2) figure out proportions

0. 500M-x x + x
   3) Kb = x² = 1.8x10^-5 *0.500 = x² = 3x10^-3 = ( conc. of OH)
   0. 500M
   4) - log(Kb) = 2.53 ( approx 3)
   5) 2.53+ pH = 14, pH = 11.47

Question 17

Calculate the K_a of methylammonium ion, where K_b of methylamine is 3.7x10^-4

Answer 17

1) We know that methylammonium and methylamine are a conjigate acid/ base. Therefore (K_a)(K_b) = K_w Where K_w= 1.0x10^-14
2) solve fore K_a

= 2.7x10^-11

Question 18

derive pKa + pKb = pKw

Answer 18

Ka + Kb = Kw

Log(Ka+Kb) = Log(Kw)

Log(Ka)+log(kb) = log(kw)

(-log(ka))+(log(kb))=-log(Kw)

-log(ka)=pKa, -log(Kb) = pkb, -log(Kw) = pKw

therefore:

pKa + pKb= Pkw

Question 19

What is the pH from the following in water;

1) Strong acid + weak base
2) weak acid + strong base

Answer 19

1) pH > 7
2) pH < 7

Question 20

Calculate the pH or a .25 M solution of CH₃COONa

Ka = 1.8x10^-5

Answer 20

pH= 9

1) Write the equaiton:

CH₃COONa + H₂O = CH₃COO^-+ Na

Then:

CH<sub>3</sub>COO<sup>-</sup> + H<sub>2</sub>O CH<sub>3</sub>COOH + H<sub>2</sub>O

I .25 m 0 0

C .-x +x +x

E 0.25-x = x + x

Kb = x/ .25

2) use (Kb)(Ka) = Kw

(x/0.25 )(1.8x10^-5) = 1.00 x 10^-14

x = 1.2 x 10^{-5 ,} x is the concentration of OH. therefore,

3) take - log of OH to get the pOH= 5.00

pH + 5.0=14,

pH= 9

Question 21

Calculate the pH of a solution that is 0.15 M NH₃ (K_b =1.8x10^-5) and 0.35 M NH₄NO₃.

Answer 21

9

1) Write the chemical equation:

NH₃ + H₂O NH₄ + OH^-

 INITIAL         I            0.15                         0.35     0

 CHANGE     C           -X                            +X      +X

EQUATION E 0.15-X 0.35+X +X

K_{b =} 1.8x10^{-5 =} (0.35 -x) (x) / (0.15-x) *****since x is so small, disregard. example 1-0.0000001 is so close to 1 it rounds up to 1 with significant figs.

2) solve for Kb = 7.7 x 10^-6
3) since Kb is only in refernce to the OH, take -log of Kb = 5.11, therefore,
4) 5.+ pH = 14 = 9.

Question 22

1) What do buffer solutions do?

How do buffer solutions work?

Answer 22

1) they resist changes in pH.
2) they work by removing protons and hydroxide from a solution.

Question 23

henderson halk.. eq.

equation used for buffer calculations

Answer 23

pH= pka + log (A-/HA)

Question 24

What is the pH of the buffer solution that is 0.24 M NH₃ AND 0.20 m NH₄Cl?

ka = 5.6x10^-10

Answer 24

pH= 9.33

1) Identify which is the acid and which is the base. NH₃ is the acid
2) use pH=pka + log(-A/HA)

pH = -log(Ka) + log (NH₄Cl/NH₃)

pH= 9.33

show your work:

pH = 9.25 + log(0.24/0.20)