Asked by Rinzin
To make a solution with a pH = 4.40 a student used the following procedure :
a certain amount of sodium acetate along with 0.280 moles of acetic acid is added to enough water to make a solution of 1.00 L
How many grams of sodium acetate were added ?
Ka = 1.80 x 10 -5
a certain amount of sodium acetate along with 0.280 moles of acetic acid is added to enough water to make a solution of 1.00 L
How many grams of sodium acetate were added ?
Ka = 1.80 x 10 -5
Answers
Answered by
DrBob222
Use the Henderson-Hasselbalch equation.
pH = pKa + log (base/acid)
pH = pKa + log (base/acid)
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