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The first and second dissociation constants of an acid H_{2}A\: are\: 1.0\times 10^{-5}and\: 5.0\times 10^{-10} respectively. The overall dissociation constant of the acid will be

Option: 1

0.2\times 10^{5}


Option: 2

5.0\times 10^{-5}


Option: 3

5.0\times 10^{15}


Option: 4

5.0\times 10^{-15}


Answers (1)

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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

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H_{2}A\rightleftharpoons H^{+}+HA^{-};

K_{1}=\frac{\left [ H^{+} \right ]\left [ HA^{-} \right ]}{\left [ H_{2}A \right ]}=1\times 10^{-5}

HA\rightleftharpoons H^{+}+A^{2-};\; K_{2}=5\times 10^{-10}=\frac{\left [ H^{+} \right ]\left [ A^{2-} \right ]}{\left [ HA^{-} \right ]}

K=\frac{\left [ H^{+} \right ]^{2}\left [ A^{2-} \right ]}{\left [ H_{2}A \right ]}=K_{1}\times K_{2}=1\times 10^{-5}\times 5\times 10^{-10}

=5\times 10^{-15}

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manish

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