[OH^-] pH 0.045 M NaCH3COO 0.055 M Ba(CH3COO)2 mixture

[OH^-] pH 0.045 M NaCH3COO 0.055 M Ba(CH3COO)2 mixture

 

 

Calculate the [OH^-] and pH for a mixture that is

0.045 M in NaCH3COO and 0.055 M in Ba(CH3COO)2.

Ka of CH3COOH is 1.8×10^(-5).

 

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Kb = Kw / Ka

= (10^(-14)) / (1.8×10^(-5))

= 5.6×10^(-10)

 

 

 

0.045 M in NaCH3COO

NaCH3COO(aq) → Na^+(aq) + CH3COO^-(aq)

[NaCH3COO] = [CH3COO^-] = 0.045 M

 

 

 

x = [Kb × C]^(1/2)

( 참고 https://ywpop.tistory.com/4294 )

 

= [5.6×10^(-10) × 0.045]^(1/2)

= 5.02×10^(-6) M = [OH^-]

 

 

 

0.055 M in Ba(CH3COO)2

Ba(CH3COO)2(aq) → Ba^2+(aq) + 2CH3COO^-(aq)

[CH3COO^-] = 2[Ba(CH3COO)2] = 2 × 0.055 M = 0.11 M

 

 

 

x = [Kb × C]^(1/2)

= [5.6×10^(-10) × 0.11]^(1/2)

= 7.85×10^(-6) M = [OH^-]

 

 

 

5.02×10^(-6) + 7.85×10^(-6) = 1.29×10^(-5) M

---> [OH^-] for a mixture

 

 

 

pOH = –log(1.29×10^(-5)) = 4.89

 

pH = 14 – 4.89 = 9.11

---> pH for a mixture