Determine of the Dissociation Constant of a Weak Acid The pH at one-half the equ
ID: 495711 • Letter: D
Question
Determine of the Dissociation Constant of a Weak Acid The pH at one-half the equivalence point in an acid-base titration was found to be 5.67. What is the value of K_a for this unknown acid? If 30.15 mL of 0.0995 M NaOH is required to neutralize 0.279 g of an unknown acid, HA, what is the molar mass of the unknown acid? Assuming that K_a is 1.85 times 10^-5 for acetic acid, calculates the pH at one-half the equivalence point and at the equivalence point for a titration of 50 mL of 1.00 M acetic acid with 0.100 M NaOH.Explanation / Answer
7) At half equivalence point [conjugate base] = [acid]
The pH of the solution (buffer) is calculated by
pH = pKa + log [conjugate base] /[acid]
5.67 = pKa + log 1.0
Thus pKa = 5.67
and ka = 2.138x 10-6
8 ) For neutralization
moles of acid = moles of base
0.279g / molar mass =30.15x10-3 L x 0.0995M
Thus molar mass = 93.00g/mol
9) CH3COOH + NaOH -------------------> CH3COONa + H2O
50x0.10 0 0 - initial mmoles
25x0.1 change
2.5 0 2.5 at half equivalence point
Thus pH = pKa + log [conjugate base]/[acid]
= (5-log 1.8) + log 2.5/2.5
= 4.75
At equivalence point
CH3COOH + NaOH -------------------> CH3COONa + H2O
0 0 5 -
thus the solution has only salt of strong base and weak acid whose pH is calculated as
pH = 1/2 [pkw +pKa + log C]
= 1/2 [ 14 + 4.74 + log (5/100)]
= 8.7194
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