Aspirin is a weak acid with a Ka of 3.0X10-5. Calculate the pH of a solution by

Question

Aspirin is a weak acid with a Ka of 3.0X10-5. Calculate the pH of a solution by dissolving 0.65g of aspirin in water and diluting it to 50.0ml. You may use the symbol RCOOH to represent aspirin when writing chemical reaction. Molecular weight of aspirin = 180.0g/mol Does the pH value that you calculate for the aspirin solution make sense based on the Ka value? Calculate the pH of a solution prepared by mixing 20.00ml of the aspirin solution in the question above with 5.00ml of 0.200M NaOH. Calculate the pH of a solution prepared by mixing 20.00ml of the aspirin solution in the question above with 10.00ml of 0.200M NaOH.

Explanation / Answer

Moles of Aspirin = mass/molar mass= 0.65g/180 = 3.6x10-3

Molarity = moles /volume = (3.6x10-3 )x1000/50 = 0.072 M

pKa = -log(ka) = 4.52

pH = 1/2pKa -1/2logc = 1/2x4.52 - 1/2log(0.072) = 2.78

-----------------------------------------------

RCOOH      +    NaOH --------->   RCOONa    + H20

1.44 millimoles       1 millmoles     

after reaction, its a buffer solution

RCOOH = 0.44millimoles, Salt RCOONa = 1

pH = pKa   + log(salt/acid) = 4.52 + log(1/0.44) = 4.87

---------------------------------------

RCOOH      +    NaOH --------->   RCOONa    + H20

1.44                 2

All acid is consumed,

The base left = 2-1.44 = 0.56 millimole

[OH-] = 0.56/30 = 0.0186

pH = 14 - pOH = 14 - (-log(0.0186) = 12.27

its a basic solution

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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