CH 8 Flashcards
Buffer action
the resistance to a change in pH
a combination of weak acid and conj base or weak base and conj acid
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
EXAMPLE OF BUFFER ACTION
acetic acid (WA) and sodium acetate(salt)
- buffer system
if 1 mL of 0.1 N of HCL is added to water what happens to pH
the pH decreases from 7 to 3
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
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
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.
sodium acetate in water
NaCH3COO- —> Na+ +ch3coo-
Acetic acid in water
HAc +H2O = H30+ Ac-
sodium acetate and acetic acid
have the same dissociation constant equation and value
ka= 1.75 x 10^-5
dissociation of sodium acetate
Na+CH3COO- —-> Na+ + CH3COO-
CH3COO- + H20 = OH- + CH3COOH
Kb= [OH-] [CH3COOH] / [CH3COO-]
pH= pKa + log [salt]/[acid]
pH= pKa + log [Ac-]/[HAc]
[OH-] =Kb (base/salt)
[H30+]= Ka (acid/salt)
pH= Pkw + pKb + log [base]/[salt]
mole ratio
salt /acid= 1.74/1
mole percent
salt/salt +acid x 100%
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
