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

1 Answer
1s2s2p
Oct 18, 2017

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

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