How do I make a buffer with a target pH of 4.8? All the solutions I have are 0.1

Question

How do I make a buffer with a target pH of 4.8?

All the solutions I have are

0.1 M acetic acid

0.1 M citric acid

0.1 M NaH2PO4

0.2 M NaOH

0.2 M HCL

Solids

Sodium acetate trihydrate (MM=136 g/mol)

Sodium hydrogen phosphate heptahydrate (MM=269 g/mol

Sodium dihydrrogen citrate (MM=211g/mol)

Aqueous equilibrium constants

acetic acid: ka = 1.8*10^-5

citric acid: ka1= 7.4*10^-4

H2PO4-: ka2 = 6.2*10^-8

a) I need to show the calculations to find how much acid (given solutions with the specifc molarity) and conjugate base (given solid salts) is necessary to make 25ml of a buffer with a pH of 4.8.

b) The buffer needs to have enough capacity to maintain its pH +/-1.0 pH units when 10ml of 0.2 M NaOH is added

c) The buffer needs to have enough capacity to maintain its pH +/-1.0 pH units when 10ml of 0.2 M HCl is added

Explanation / Answer

a) pH of buffer=4.8

Using henderson-hasselbach equation,

pH=pka+log [conjugate base]/[acid]

for buffer prepared using acetic acid, use ka=1.8*10^-5

pka=-log ka=-log (1.8*10^-5 )=4.745

So,log [conjugate base]/[acid]=pka-pH=4.745-4.8=-0.055

[conjugate base]/[acid]=10^-0.055=0.881

As volume of buffer=0.025L so mol of acetic acid added=0.1mol/L*0.025L=0.0025 mol

Thus mol of conjugate base required=0.881*0.0025mol=0.0022 mol

mass of conjugate base (Sodium acetate trihydrate),=mol*molar mass=0.0022mol*136g/mol=0.299 g

So,0.299 g of (Sodium acetate trihydrate is required to be added to 25 ml of 0.1M acetic acid to prepare a buffer of pH=4.8

b)when 10ml of 0.2 M NaOH is added,

mol of base added=0.010L*0.2mol/L=0.002 mol

acetic acid+NaOH---> sodium acetate +H2O

So, 0.002 mol of acetic acid is converted to sodium acetate

Thus, mol of acetic acid remaining in the solution=0.0025mol-0.002mol=0.0005mol

mol of sodium acetate(conjugate base ) in the solution=0.0022mol+0.002 mol=0.0042 mol

thus ratio of [conjugate base]/[acid]=0.0042/0.0005=8.4

pH''=pka+log [conjugate base]/[acid]=4.745+log 8.4=5.669=5.7-4.8

thus change in pH=5.7-4.8=0.9 <1.0

pH maintained to +/-1.0 pH units

b)

Similarly,when 10ml of 0.2 M HCl is added

mol of base added=0.010L*0.2mol/L=0.002 mol

sodium acetate+HCl---> acetic acid +H2O

So, 0.002 mol of sodium acetate is converted to acetic acid

Thus, mol of acetic acid remaining in the solution=0.0025mol+0.002mol=0.0045mol

mol of sodium acetate(conjugate base ) in the solution=0.0022mol-0.002 mol=0.0002 mol

thus ratio of [conjugate base]/[acid]=0.0002/0.0045=0.0444

pH''=pka+log [conjugate base]/[acid]=4.745+log 0.04=3.5

thus change in pH=4.8-3.5=1.3 (slightly deviated)

pH maintained to +/-1.0 pH units

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.