Question

What is the pH of a 0.150 M solution of sodium acetate (NaO2CCH3)? Ka(CH3CO2H) = 1.8 x 10-5.

Answer 1

The pH is roughly 8.96.

Explanation:

Sodium acetate is the salt of a weak acid and strong base from the equation:
`C_(2)H_(3)NaO_2->CH_3COO^(-)+Na^(+)`, where: `CH_3COO^(-)+H_2O\\rightleftharpoonsCH_3COOH+OH^(-)`

As it is a weak acid and strong base, this is a good indicator of a fairly high pH.

`K_b=([HB^+][OH^-])/([B])` where:

• `[B]` is the concentration of the base
• `[HB^+]` is the concentration of base ions.
• `[OH^-]` is the concentration of the hydroxide ions.

Also, `K_aK_b=1*10^(-14)`/ So, `K_b=(1*10^(-14))/(1.8*10^(-5))=5.555...*10^(-10)`

`[("",CH_3COO^(-),CH_3COOH,OH^-),(I,0.150,0,0),(C,-x,+x,+x),(E,0.150-x,x,x)]`

`K_b=5.555...*10^(-10)=x^2/(0.150-x)`

Since the value for `K_b` is small, we will assume the the value for `x` is small, and so will take `0.150`.

So, `x=sqrt(0.150(5.555...*10^(-10)))=OH^-`

`pH=14-pOH=14-(-log(OH^-))`
`=14-(-log(sqrt(0.150(5.555...*10^(-10)))))`
`=14-(-log(9.128709292*10^(-6)))`
`=14-5.039590623~~8.96`