CH 8 Flashcards by c v

Buffer action

the resistance to a change in pH

a combination of weak acid and conj base or weak base and conj acid

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Buffer

is a compound or mixture of compounds, by their presence in a solution, resist changes in pH upon addition of small quantities of acid or alkali

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EXAMPLE OF BUFFER ACTION

acetic acid (WA) and sodium acetate(salt)
- buffer system

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if 1 mL of 0.1 N of HCL is added to water what happens to pH

the pH decreases from 7 to 3

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if the strong acid, hcl IS ADDED TO 0.01m solution of sodium acetate and acetic acid, what happens to pH

the pH only changes by 0.09 ph units because the base(sodium acetate) Ac- ties up the hydrogen ions

Ac- +h30+ = HAc +H20

strong acid(HCL) is neutralized by weak base (sodium acetate) - neutralizes the H+ ions

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if a strong base, NaOH (sodium hydroxide) is added to the buffer mixture, acetic acid neutralizes the hydroxide ions

strong base (NaOH) is neutralized by weak acid (acetic acid) -neutralizes the OH- ions

HAc + OH- = H20 + Ac-

strong acid(HCL) is neutralized by weak base (sodium acetate) - neutralizes the H+ ions

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common ion effect

Salt uses a common ion to incomplete solubility of that weak acid or base

The common ion effect refers to the phenomenon in which the addition of an ion that is common to the solute in a solution decreases the solubility of the solute. This effect occurs when a compound with a particular ion is added to a solution already containing that ion.

In the case of acetic acid (CH3COOH) and sodium acetate (CH3COONa), both compounds contain the acetate ion (CH3COO^-). When sodium acetate, which dissociates into sodium ions (Na^+) and acetate ions (CH3COO^-), is added to a solution containing acetic acid, the acetate ion is the common ion.

CH3COOH ⇌ CH3COO^- + H^+

Now, if you add sodium acetate to this solution, the acetate ions from sodium acetate will be in equilibrium with the acetate ions already present from the acetic acid. According to Le Chatelier’s principle, the increase in acetate ion concentration will shift the equilibrium to the left (towards reactants), reducing the dissociation of acetic acid:

CH3COOH + CH3COO^- ⇌ CH3COO^- + H^+

As a result, more undissociated acetic acid will be formed, and the overall concentration of acetate ions in the solution will increase while the concentration of hydrogen ions decreases. This leads to a decrease in the ionization of acetic acid and a decrease in its acidity.

The common ion effect ultimately reduces the ionization of a weak acid or a weak base when a salt of that acid or base is added to its solution, causing a shift in the equilibrium towards the formation of the undissociated molecules.

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sodium acetate in water

NaCH3COO- —> Na+ +ch3coo-

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Acetic acid in water

HAc +H2O = H30+ Ac-

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sodium acetate and acetic acid

have the same dissociation constant equation and value

ka= 1.75 x 10^-5

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dissociation of sodium acetate

Na+CH3COO- —-> Na+ + CH3COO-

CH3COO- + H20 = OH- + CH3COOH

Kb= [OH-] [CH3COOH] / [CH3COO-]

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pH= pKa + log [salt]/[acid]

pH= pKa + log [Ac-]/[HAc]

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[OH-] =Kb (base/salt)

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[H30+]= Ka (acid/salt)

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pH= Pkw + pKb + log [base]/[salt]

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mole ratio

salt /acid= 1.74/1

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mole percent

salt/salt +acid x 100%

free base %

free base% -100= salt form or unionized form

buffer capacity (B)

the magnitude of resistance of a buffer to pH changes

Buffer efficiency; buffer index and buffer value

B= change B/change pH

CHANGE b: Small increments in gram eq. per liter of strong base added to the buffer to produce pH change

it depends on the values of the ratio of salt/acid

buffer capacity depends on magnitude of the individual concentrations of the buffer components

acetic acid and sodium acetate each 0.1M in water

pH=4.76 + log (0.1+0)/(0.1-0)=4.76

pH= pKa +log (salt +base)/ (salt-base)

at a ph =4.76

the moles of NaOH is 0 and buffer capacity is 0

if the pH of a solution of sodium acetate and acetic acid increases what happens to buffer capacity and the moles of NaOH added?

summary, as the pH of a solution containing acetic acid and sodium acetate increases, the buffer capacity decreases due to the shift in equilibrium between the weak acid and its conjugate base toward the dominant presence of the conjugate base at higher pH levels.

the buffer capacity decrease with increase in pH

the moles of NaOH added will increase with increase of pH

B= 2.3c (Ka[h30+]/ (ka + [h30+])^2

c= the total buffer concentrations
the sum of the molar concentration of acid and salt

preparation of pharmaceutical buffer solution

select a weak acid having a pKa close to pH from the buffer eq calculate the ratio of salt and weak acid (total concentration c= acid + salt) consider the individual concentrations of buffer salt and acid: concentration: 0.05=0.5M buff capacity: 0.01= 0.1M DETERMINE the pH and buffer capacity using the pH meter

max buffer capacity

Bmax = 2.3C (h30+0^2/2(H3O+)^2= 2.303/4c OR Bmax= 0.576C

neutralization curve and buffer capacity

neutralization of a strong acid and a weak acid by strong base acetic acid (wa) and naOH(sb)= ph =6 hcl (sa) and naOh(sb)= ph= 2

neutralization of a tribasic acid

3 stages for H3PO4 H3PO4= PK1=2.21 H2PO4-= PK2=7.21 HPO4 2-=PK3= 12.67

BLOOD

Ph =7.4

buffers in biological and pharma systems

in vivo biologic buffer systems -blood -lacrimal fluid (tear) ph 7.4(7-8) urine ph=6.0 (4.5-7) when the urine pH is lower than normal hydrogen ions are excreted by kidneys when urine pH is above 7.4, hydrogen ions are retained by kidneys to return it to normal pH pharma buffer: -ophthalmic and parenteral solution these need to be prepared so the pH in both doesn't change and affect the abrade surfaces and tissue

pharmaceutical solns that are meant to be applied need to have same osmotic pressure as the body fluids

isotonic solutions

have the same osmotic pressure of solns to the human body . same osmotic pressure as the RBC and is isotonic with blood no swelling or contraction of tissue in contact with no discomfort in eye, nasal, blood,etc example: isotonic sodium chloride NaCl (0.9%)

hypertonic

RBC in a 2% NaCl soln RBC is shrink and crenated, water leaves the cell or the NaCl leaves cell

hypotonic

0.2% of Nacl in RBC content hemolysis of the RBC, water enter cells and causes it to burst

adjusting tonicity

class 1 methods and class 2 methods

class 1 methods

in class one method sodium chloride or some other substance is added to soln of drug LOWER temp soln to -0.52C, thus isotonic in body cryoscopic method sodium chloride eq method

class 2 methods

class 2 water is added to drug to make it isotonic, buffered isotonic dilution solution white vincient method (isotonic soln) sprowl methods

0.58 degrees C

is freezing pt of 1% of NaCl

0.9% NaCl freezing pt

0.52 DEGREES C

EQUATION : TEMP OF NACl/ percent soln= x/0.9%

ethanolamine

base