Let the volume of 1 N acetic acid, be V.
acetic acid Ka = 1.8 x 10^-5
[ H+ ] = sqrt(Ka x c)
We need concentration c .
3.5mL of 1N CH3COOH is added.
So Number of normals = 3.5 x 1 milli
Net vol = 38.5 ml
Normality = 3.5 * 1 /38.5 = 0.091 N
[ H+ ] = sqrt(Ka x c)
pH = -log( sqrt(Ka *c) ) = 2.89
pH = 2.89
Assume that you have collected 35 ml egg white. Answer the following question: What will be...
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3.5 mL of 1 N acetic acid is added to 35 mL of a neutral solution. What is the resulting pH of the solution?
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