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Can someone answer this acid/base question?
A solution is prepared by adding 0.002 moles of sodium acetate to a 100 mL of 0.03 M acetic
acid.
(Ka (CH3COOH) = 1.8 x 10-5
)
a) Assuming that no volume change occurs upon addition of the sodium acetate what is the
pH of the solution?
1 Answer
- 6 years ago
Sodium Acetate = CH3COONa
Acetic Acid = CH3COOH
In this case because sodium acetate is the limiting reactant, only 0.002 moles will of products will be produced. Aka only 0.002 moles of H+ will be produced.
pH = 1/2 (pKa - log[H+])
pH = 1/2 ( (-log(1.8x10^-5)) - (log[0.002]) )
pH = 1/2 (7.443)
pH = 3.72
