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A 0.1276−g sample of a monoprotic acid (molar mass = 1.10 × 102 g/mol) was dissolved...

A 0.1276−g sample of a monoprotic acid (molar mass = 1.10 × 102 g/mol) was dissolved in 25.0 mL of water and titrated with 0.0633 M NaOH. After 10.0 mL of base had been added, the pH was determined to be 4.87. What is the Ka for the acid? Answer in scientific notation.

Expert Solution

The number of moles of acid

The number of moles of NaOH(OH-) in 10.0mL of solution is

The neutralization reaction is

HA(aq) + OH^- (aq) A^-(aq) + H2O (l)

Initial (mol) 0.00116 6.33 x 10^-4 0

Change (mol) -6.33 x 10^-4 -6.33 x 10^-4 +6.33 x 10^-4

Final (mol) 5.27 x 10^-4 0 +6.33 x 10^-4

Now the weak acid equilibrium will be reestablised. Since both HA and A,

by using Henderson–Hasselbalch equation we can calculate Ka

queen_honey_blossom answered 2 years ago

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