Calculate the pH of a 0.30 M NaCH 3 COO (sodium acetate) solution. a. 4.9 b. 7.0

Question

Calculate the pH of a 0.30 M NaCH3COO (sodium acetate) solution. a. 4.9 b. 7.0 c. 9.1 d. 9.8

I am confused on the part where you have to find the "Ka" value for "HNa" I look through my book and even online and I
could not find the Ka value for "HNa" Do you even look for the value for "HNa"? I even looked to see if "HNa" was a
strong acid and because maybe it was 100% dissociable but did not find that either. Can someone please help me and
elaborate on all points, thanks and have a great day!













Calculate the pH of a 0.30 M NaCH3COO (sodium acetate) solution. Calculate the pH of a 0.30 M NaCH3COO (sodium acetate) solution. 4.9 7.0 9.1 9.8 Calculate the pH of a 0.30 M NaCH3COO (sodium acetate) solution. a. 4.9 b. 7.0 c. 9.1 d. 9.8

I am confused on the part where you have to find the "Ka" value for "HNa" I look through my book and even online and I
could not find the Ka value for "HNa" Do you even look for the value for "HNa"? I even looked to see if "HNa" was a
strong acid and because maybe it was 100% dissociable but did not find that either. Can someone please help me and
elaborate on all points, thanks and have a great day!

Explanation / Answer

Sodium Acetate a strong salt
CH3COONa => CH3COO- + Na+

Kb = Kw/Ka = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.6 x 10^-10

CH3COO- + H2O <------> CH3COO- + OH-

...0.3.........----------->........0........+......o.

..(0.3 - x) ..----------->........x........+......x

Kb = x^2/(0.3-x)

5.6 x 10^-10 = x^2/(0.3-x)

x = [OH-]=1.296 x 10^-5 M

pOH = - log (1.296 x 10^-5) = 4.887

pH = 14 - 4.887 = 9.113 <-----ANS

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