Biochemistry 1

Question 1

Ionization of Water

Answer 1

H₂O + H₂O ⇔ H₃O⁺ + OH^-

(Short hand version: H₂O⇔ H+ + OH-)

Question 2

Equation of ionization of Water

Kw= [H₃O⁺][OH^-] / [H₂O]²

Answer 2

Kw = [H⁺] [OH-] Kw = 1.0 x 10^-14 M²

pH = pOH = 14.00

Nuetral—–[H₃O⁺] = [OH^-] = 10^-7 M

Acidic—– [H₃O⁺] > [OH^-]

Basic—–[OH^-] > [H₃O⁺]

e.g. Household Amonia [OH-] = 5 x 10^-4 M find [H+]: [H+] = Kw / [OH-]

[H+] = 1.0 x 10^-14 M² / 5 x 10^-4 M

   = **2 x 10<sup>-11</sup> M**

Question 3

_p- function _

What equation is used to fing pH of [H+] ?

Answer 3

p= -log ₁₀

pH= -log [H₃O⁺] or (-log [H⁺]) <– same

pH + pOH = pKw = 14.00

e.g. pH of the blood :

pH= -log 4 x 10^-8

pH = 7.4

Question 4

_p-function _

Find M of [H3O+] in solution given the pH is 2.3

Answer 4

[H₃O⁺] = 10^-pH

e.g. if pH = 2.3:

[H₃O⁺] = 10 ^{-2.3 <em><span>(entered into calc)</span></em>}

= 5 x 10^-3M

Question 5

**What is the pH of a solution with: **

[H⁺] = 10^-5 M?

Answer 5

pH= -log 10 ^-5 M (entered into calc)

pH = ** 5**

Question 6

What is the pH of a solution with:

[H⁺] = 2.5 x 10^-9 M?

Answer 6

pH = -log (2.5 x 10^-9) M (entered into calc)

pH = 8.6021

pH = 8.60

Question 7

Find the [H+] and [OH-] for a solution with pH of 7.50

Answer 7

[H+] = 10^-pH

= 10^-7.50 *(entered into calc) *= 3.1623 x 10^-8 M , 3.20 x 10^-8 M

Kw = [H+][OH^-] = 1.0 x 10^-14 M²

[OH^-] = Kw/ [H⁺] = 1.0 x 10^-14 M² / 3.2 x 10^-8 M (entered into calc) = **3.125 x 10 ^-7 M **

Question 8

What donates a H⁺ proton and accepts an e^- pair?

Answer 8

Acids:

e.g.

HCl + H₂O —> H₃O⁺+ Cl^-

Question 9

What accepts a H⁺ proton and donates an e^- pair?

Answer 9

Bases:

e.g.

NH₃ + H₂O —-> NH₄⁺ + OH^-

Question 10

Do Monoprotic Acids donate a proton?

Answer 10

Yes, but only one.

Question 11

Monoprotic Acids:

What are the Strong Acids and do they dissociate?

Answer 11

Are: HNO_3, HCl, HClO₄,

H₂SO₄ —> H⁺ + HSO₄^-

• 100 % dissociation
• Co (concentration) = [H⁺]

Question 12

Monoprotic Acids

Weak Acids

Answer 12

Most biological acids are weak and —<100% dissociation.

Dissociation Constant = Ka, to find [H+] and to determine the amount of dissociation

Question 13

Monoprotic Acids

Weak Acids:

What is the short version of the Equillibrium(Eq) expression for Weak Acids?

Answer 13

HA + H₂O → H₃O⁺ + A^-

Reactants → Products

Kaq += [H₃O⁺][A^-] / [HA][H₂O]

Products / Reactants

Ka = [H₃O⁺][A^-] /[HA]

Ka = [H⁺][A^-] / [HA]

Question 14

What is the pH of a weak acid solution?

Answer 14

Use the expression for Ka:

Ka = [H⁺]² / Co - [H⁺]

(can drop the [H⁺] if [H₃0⁺] / Co <.05)

Question 15

What is the pH of a 0.10 M acetic acid solution where

pKa = 4.75?

Answer 15

pKa = 4.75 (given)

Ka = 10^-4.75 (entered into the calc) = 1.8 x 10^-5 M

Ka = [H⁺]² / Co

 =1.8 x 10<sup> -5</sup> M = [H<sup>+</sup>]<sup>2</sup> / 0.10 M *(Co given)*

[H⁺]² = Ka x Co

[H⁺] = (Ka x Co)^½= (1.8 x 10^-5 M x 0.10 M )^½ (entered into calc)= 1.3 x 10^-3 M

pH = -log 1.3 x 10^-3 M (entered into calc)= 2.88

Question 16

Calculate the pH of solutions with the following [H+] s.

A. 5 X 10^-9 M

B. 2.0 x 10^-2 M

C. .000000000083 M

D. 7.5 x 10^-6 M

Answer 16

pH = -log [H⁺]

A. -log (5 x 10^-9) (entered into calc) = 8.3 M

B. -log (2.0 x 10^-2) = 1.69 M

C. .000000000083 *(entered into calc) *= 8.3 x 10^-11

   = -log(8.3 x 10<sup>-11</sup>) (entered into calc) = **10.08 M**

D. -log (7.5 x 10^-6) (entered into calc) = 5.12 M

Question 17

Given solutions with the following pHs, determine [H₃O⁺] (same as [H⁺]) and [OH^-]:

A. 1.25

B. 10.5

C. -.50

D. 7.85

Answer 17

Kw = 1.0 x 10^-14 M²

[H⁺] = 10 ^-pH

Kw = [H⁺][OH^-]

A .[H+] = 10^-1.25 (entered into calc) = 5.6 x10^-2 M

Kw = [H⁺][OH^-]

[OH^-] = Kw / [H⁺]

= 1.0 x 10^-14 M² / 5.6 x 10^-2 M = 1.8 x 10^-13 M

repeat for BCD.

Question 18

What is the pH of a solution made from adding a weak acid to water at an original concentration of 0.55 M?

This weak acid has a pKa = 3.30

Answer 18

pKa = Ka

Ka = 10^-3.30* (entered into calc) *= 5.0 x 10^-4 M

Ka = [H+]² / Co

[H⁺]² = (Ka x Co)

= (5.0 x 10^-4 x 0.55 M)^½ (entered into calc)= 1.66 x 10^-2 M

pH = -log [H+]

  = -log (1.66 x 10<sup>-2</sup>) M *(entered into calc)* = **1.78**

Question 19

Calculate the pKa (-COOH) from your initial pH value of 1.9,

a Co = 0.10 M and the following equation:

Ka (-COOH) = [H⁺]i ² / Co - [H⁺]i

Answer 19

pKa= [H+]

Co= 0.10 M

pKa = 10^-1.90 *(entered into calc) *= 1.259 x 10^-2

Ka (-COOH) = [1.259 x 10^-2]i² / (0.10 M - [1.259 x 10 ^-2]i) (entered into calc) = 1.813 x 10^-3

pKa = -log (1.813 x 10^-3) *(entered into calc) * = 2.74