Question #6521

if a weak acid, HA, is 3% dissociated in a 0.25M solution, calculate the Ka and the pH of the solution.

Expert's answer

a=3% (0.03)

C=0.25M

Ka-? pH-?

HA <=>H+ +

A-

a=(K/C) ^ (1/2)

a^ 2= K/C

K=(a^ 2)/C

K= (0.03^ 2)

/0.25=0.0036

[H+]=(K*C) ^ (1/2)=0.03

pH=-log[H+]=1.52

K=0.0036

pH=1.52

C=0.25M

Ka-? pH-?

HA <=>H+ +

A-

a=(K/C) ^ (1/2)

a^ 2= K/C

K=(a^ 2)/C

K= (0.03^ 2)

/0.25=0.0036

[H+]=(K*C) ^ (1/2)=0.03

pH=-log[H+]=1.52

K=0.0036

pH=1.52

## Comments

Anthony28.01.13, 02:59The % dissociated (or value "a") tells you that you have 3% of A- and 97% of HA. For a 0.25M solution, that means you have (0.97 * 0.25)M HA present, (0.03 * 0.25)M A- present, and (0.03 * 0.25)M H+ present. Putting these concentrations into the equation doesn't give you K = 0.0036.

In fact, the first problem is that whoever answered before assumes "% dissociated" is the concentration of H+ present...a mistake which can be seen in solving the second part.

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