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Equilibrium

Question
CBSEENCH11008361

What is [H+] in mol/L of a solution that is 0.20 M in CH3COONa and 0.10 M in CH3COOH? 
(Ka for CH3COOH = 1.8 x 10-5 )

  • 3.5 x 10-4

  • 1.1 x 10-5

  • 1.8 x 10-5

  • 9.0 x 10-6

Solution

D.

9.0 x 10-6

CH3COOH (weaK acid) and CH3COONa (conjugated salt) form acidic buffer and for acidic buffer,
pH space equals space pK subscript straight a space plus space log space fraction numerator open square brackets salt close square brackets over denominator left square bracket acid right square bracket end fraction
and
left square bracket straight H to the power of plus right square bracket space equals space minus space antilog space pH
pH space equals space minus space log space straight K subscript straight a space plus space log space fraction numerator left square bracket salt right square bracket over denominator left square bracket acid right square bracket end fraction space space space space space left square bracket therefore space pK subscript straight a space equals space minus space log space straight K subscript straight a right square bracket
space equals space minus log space left parenthesis 1.8 space straight x space 10 to the power of negative 5 end exponent right parenthesis space plus space log space fraction numerator 0.20 over denominator 0.10 end fraction
equals space 4.74 space plus space log space 2
equals space 4.74 space plus space 0.3010 space equals space 5.041
Now comma space left square bracket straight H to the power of plus right square bracket space equals antilog space left parenthesis negative 5.045 right parenthesis
equals space 9.0 space straight x space 10 to the power of negative 6 end exponent space mol divided by straight L