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  • What is the concentration of acetic acid which can be added to 0.4 M formic acid solution so that the dissociation for both acids is the same.  Given Ka for acetic acid = 1.8*10^{-5}

What is the concentration of acetic acid which can be added to 0.4 M formic acid solution so that the dissociation for both acids is the same.  Given Ka for acetic acid = 1.8*10^{-5} and Ka for formic acid = 2*10^{-4}

 

Option: 1

5.55


Option: 2

3.33


Option: 3

6.66


Option: 4

4.44


Answers (1)

best_answer

As we have learnt in

 

 

pH of Solutions: Weak Acids -

pH of Weak Acids
Weak acids are those acids which dissociate partially in solutions. For example:

  • 8 M HA (Ka =2 x 10-8)
    The chemical equation for the dissociation of weak acid HA is as follows:
               \mathrm{HA\: \rightleftharpoons \: H^{+}\: +\: A^{-}}
    Initial:   8M            0           0
    Equil:  8 - 8?         8?         8?

    The equilibrium constant Ka for the weak acid is given as follows:
    \\\mathrm{K_{a}\: =\: \frac{[H^{+}][A^{-}]}{[HA]}\: =\: \frac{8\alpha \, .\, 8\alpha }{8(1-\alpha )}\: =\: \frac{8\alpha ^{2}}{1-\alpha }}\\\\\mathrm{K_{a}\: =\: 8\alpha ^{2}\: \: \: \: \: \: \: (as\: (1-\alpha) \: \approx \: 1)}\\\\\mathrm{Thus,\: \alpha \: =\: \sqrt{\frac{K_{a}}{8}}\: =\: \sqrt{\frac{2\, x\, 10^{-8}}{8}}\: =\: \sqrt{\frac{10^{-8}}{4}}\: =\: 0.5\, x\, 10^{-4}}\\\\\mathrm{[H^{+}]\: =\: 8\,\, x\,\, 0.5\,\, x\,\, 10^{-4}\: =\: 4\: x\: 10^{-4}}\\\\\mathrm{pH\: =\: -log_{10}4\: +\: 4}\\\\\mathrm{pH\: =\: -0.60\: +\: 4\: =\: 3.4}
    Thus, pH of this given acid = 3.4
     
  • 0.002N CH3COOH(? = 0.02)
    The chemical equation for the dissociation of CH3COOH is as follows:
               \mathrm{CH_{3}COOH\: \rightleftharpoons \: CH_{3}COO^{-}\: +\: H^{+}}
    Initial:           c                              0                   0
    Equil:         c - c?                         c?                c?

    The equilibrium constant Ka for the weak acid is given as follows:
    \\\mathrm{K_{a}\: =\: \frac{[CH_{3}COOH^{-}][H^{+}]}{[CH_{3}COOH]}\: =\: \frac{c\alpha \, .\, c\alpha }{c(1-\alpha )}\: =\: \frac{c\alpha ^{2}}{1-\alpha }}\\\\\mathrm{K_{a}\: =\: c\alpha ^{2}\: \: \: \: \: \: \: (as\: (1-\alpha) \: \approx \: 1)}\\\\\mathrm{Now,\: as\: we\: have\: given:}\\\mathrm{c\: =\: 0.002N\: or\: 0.002M\: \: \: \: \: \: (Normality\: =\: Molarity,\: as\: n\: factor\: =\: 1)}\\\mathrm{\alpha \: =\:\frac{2}{100}\: =\: 0.02}\\\\\mathrm{Thus,\: [H^{+}]\: =\: 0.002\,\, x\,\, 0.02\: =\: 4\: x\: 10^{-5}}\\\\\mathrm{pH\: =\: -log_{10}(4\: x\: 10^{-5})}\\\\\mathrm{pH\: =\: 5\: -\:log 4\: =\: 4.4}
    Thus, pH of acetic acid = 4.4

-

 

 

As it is given that both the acids are present in the mixture,

hence [H^{+}] _{HCOOH} = [H^{+}] _{CH_{3}COOH}

\Rightarrow c_{1}\alpha _{1} = c_{2}\alpha _{2}

Or \sqrt{K_{a1}}c_{1} = \sqrt{K_{a2}}c_{2}

\Rightarrow {K_{a1}}c_{1} = {K_{a2}}c_{2}

0.4*2*10^{-4} = c*1.8*10^{-5}

=> c = 4.44

Posted by

Rishi

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