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Acids and bases, pH and buffers. Dr. Mamoun Ahram Lecture 2. Acids and bases. Acids versus bases. Acid: a substance that produces H+ when dissolved in water (e.g., HCl, H2SO4) Base: a substance that produces OH- when dissolved in water (NaOH, KOH) What about ammonia (NH3)?.

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Acids and bases, pH and buffers


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acids and bases ph and buffers

Acids and bases, pH and buffers

Dr. Mamoun Ahram

Lecture 2

acids versus bases
Acids versus bases
  • Acid: a substance that produces H+ when dissolved in water (e.g., HCl, H2SO4)
  • Base: a substance that produces OH- when dissolved in water (NaOH, KOH)
  • What about ammonia (NH3)?
br nsted lowry acids and bases
Brønsted-Lowry acids and bases
  • The Brønsted-Lowry acid: any substance able to give a hydrogen ion (H+-a proton) to another molecule
    • Monoprotic acid: HCl, HNO3, CH3COOH
    • Diprotic acid: H2SO4
    • Triprotic acid: H3PO3
  • Brønsted-Lowry base: any substance that accepts a proton (H+) from an acid
    • NaOH, NH3, KOH
acid base reactions
Acid-base reactions
  • A proton is transferred from one substance (acid) to another molecule

Ammonia (NH3) + acid (HA)  ammonium ion (NH4+) + A-

    • Ammonia is base
    • HA is acid
    • Ammonium ion (NH4+) is conjuagte acid
    • A- is conjugate base
water acid or base
Water: acid or base?
  • Both
  • Products: hydronium ion (H3O+) and hydroxide
amphoteric substances
Amphoteric substances
  • Example: water

NH3 (g) + H2O(l) ↔ NH4+(aq) + OH–(aq)

HCl(g) + H2O(l) → H3O+(aq) + Cl-(aq)

acid base reactions1
Acid-base reactions

Acid + base  salt + H2O

  • Exceptions:
  • Carbonic acid (H2CO3)-Bicarbobate ion (HCO3-)
  • Ammonia (NH3)-
slide10
Rule
  • The stronger the acid, the weaker the conjugate base

HCl(aq)→ H+(aq) + Cl-(aq)

NaOH(aq)→ Na+(aq) + OH-(aq)

HC2H3O2 (aq) ↔ H+(aq) + C2H3O2-(aq)

NH3 (aq) + H2O(l)↔ NH4+(aq) + OH-(aq)

equilibrium constant
Equilibrium constant

   HA  <-->   H+ + A-

Ka: >1 vs. <1

expression
Expression
  • Molarity (M)
  • Normality (N)
  • Equivalence (N)
molarity of solutions
Molarity of solutions

moles = grams / MW

M = moles / volume (L)

grams = M x vol (L) x MW

exercise
Exercise
  • How many grams do you need to make 5M NaCl solution in 100 ml (MW 58.4)?

grams = 58.4 x 5 moles x 0.1 liter = 29.29 g

normal solutions
Normal solutions

N= n x M (where n is an integer)

  • n =the number of donated H+

Remember!

The normality of a solution is NEVER less than the molarity

equivalents
Equivalents
  • The amount of molar mass (g) of hydrogen ions that an acid will donate
    • or a base will accept
  • 1M HCl = 1M [H+] = 1 equivalent
  • 1M H2SO4 = 2M [H+] = 2 equivalents
exercise1
Exercise
  • What is the normality of H2SO3 solution made by dissolving 6.5 g into 200 mL? (MW = 98)?
example
Example

One equivalent of Na+ = 23.1 g

One equivalent of Cl- - 35.5 g

One equivalent of Mg+2 = (24.3)/2 = 12.15 g

Howework:

Calculate milligrams of Ca+2 in blood if total concentration of Ca+2 is 5 mEq/L.

titration
Titration
  • The concentration of acids and bases can be determined by titration
excercise
Excercise
  • A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl. What was the concentration of the HCl?
  • Step 1 - Determine [OH-]
  • Step 2 - Determine the number of moles of OH-
  • Step 3 - Determine the number of moles of H+
  • Step 4 - Determine concentration of HCl
a 25 ml solution of 0 5 m naoh is titrated until neutralized into a 50 ml sample of hcl
A 25 ml solution of 0.5 M NaOH is titrated until neutralized into a 50 ml sample of HCl
  • Moles of base = Molarity x Volume
  • Moles base = moles of acid
  • Molarity of acid= moles/volume
another method
Another method

MacidVacid = MbaseVbase

slide24
Note
  • What if one mole of acid produces two moles of H+

MacidVacid = 2MbaseVbase

homework
Homework
  • If 19.1 mL of 0.118 M HCl is required to neutralize 25.00 mL of a sodium hydroxide solution, what is the molarity of the sodium hydroxide?
  • If 12.0 mL of 1.34 M NaOH is required to neutralize 25.00 mL of a sulfuric acid, H2SO4, solution, what is the molarity of the sulfuric acid?
equilibrium constant1
Equilibrium constant

Keq = 1.8 x 10-16 M

slide29
Kw
  • Kw is called the ion product for water
acid dissociation constant
Acid dissociation constant
  • Strong acid
  • Strong bases
  • Weak acid
  • Weak bases
the equation
The equation

pKa is the pH where 50% of acid is dissociated into conjugate base

problems and solutions
Problems and solutions
  • A solution of 0.1 M acetic acid and 0.2 M acetate ion. The pKa of acetic acid is 4.8. Hence, the pH of the solution is given by
  • Similarly, the pKa of an acid can be calculated
exercise2
Exercise
  • What is the pH of a buffer containing 0.1M HF and 0.1M NaF? (Ka = 3.5 x 10-4)
homework1
Homework
  • What is the pH of a solution containing 0.1M HF and 0.1M NaF, when 0.02M NaOH is added to the solution?
at the end point of the buffering capacity of a buffer it is the moles of h and oh that are equal
At the end point of the buffering capacity of a buffer, it is the moles of H+ and OH- that are equal
exercise3
Exercise
  • What is the concentration of 5 ml of acetic acid knowing that 44.5 ml of 0.1 N of NaOH are needed to reach the end of the titration of acetic acid? Also, calculate the normality of acetic acid.
excercise1
Excercise
  • What is the pH of a lactate buffer that contain 75% lactic acid and 25% lactate? (pKa = 3.86)
  • What is the pKa of a dihydrogen phosphae buffer when pH of 7.2 is obtained when 100 ml of 0.1 M NaH2PO3 is mixed with 100 ml of 0.1 M Na2HPO3?
buffers in human body
Buffers in human body
  • Carbonic acid-bicarbonate system (blood)
  • Dihydrogen phosphate-monohydrogen phosphate system (intracellular)
  • Proteins
blood buffering
Blood buffering

Blood (instantaneously)

CO2 + H20

H2CO3

H+ + HCO3-

Lungs

(within minutes)

Excretion via kidneys (hours to days)

roles of lungs and kidneys
Roles of lungs and kidneys
  • Maintaining blood is balanced by the kidneys and the lungs
  • Kidneys control blood HCO3 concentration ([HCO3])
  • Lungs control the blood CO2 concentration (PCO2)
acidosis and alkalosis
Acidosis and alkalosis
  • Can be either metabolic or respiratory
  • Acidosis:
    • Metabolic: production of ketone bodies (starvation)
    • Respiratory: pulmonary (asthma; emphysema)
  • Alkalosis:
    • Metabolic: administration of salts or acids
    • Respiratory: hyperventilation (anxiety)
acid base imbalances
Acid-Base Imbalances
  • pH< 7.35 acidosis
  • pH > 7.45 alkalosis
respiratory acidosis
Respiratory Acidosis

H+ + HCO3- H2CO3  CO2 + H2O

respiratory alkalosis
Respiratory Alkalosis

H+ + HCO3- H2CO3  CO2 + H2O

metabolic acidosis
Metabolic Acidosis

H+ + HCO3- H2CO3  CO2 + H2O

metabolic alkalosis
Metabolic Alkalosis

H++ HCO3- H2CO3  CO2 + H2O