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  • A solution contains 0.08 M HCl, 0.08 M CHCl2COOH and 0.1 M CH3COOH. The pH of this solution is 1. If Ka for acetic acid is 10-5, </

A solution contains 0.08 M HCl, 0.08 M CHCl2COOH and 0.1 M CH3COOH. The pH of this solution is 1. If Ka for acetic acid is 10-5, then find the Ka for CHCl2COOH

Option: 1

1.67\times10^{-2}


Option: 2

3.33\times10^{-2}


Option: 3

4.67\times10^{-2}


Option: 4

None of the above


Answers (1)

best_answer

The dissociation of CHCl2COOH takes place as

                               \mathrm{CHCl}_{2} \mathrm{COOH} \rightleftharpoons \mathrm{CHCl}_{2} \mathrm{COO}^{-}+\mathrm{H}^{+}

Initial concentration      0.08                                 0                          0.08 (HCl)

Final Concentration     0.08 - a                            a                           0.08+a

\therefore \quad\left[\mathrm{H}^{+}\right]=0.08+a.....................(1)

but pH is given as 1

 so, \quad\left[\mathrm{H}^{+}\right]=0.1...........................(2)

equating equation (1) and (2)

a = 0.02

Ka for CHCl2COOH  will be given as:

\begin{aligned} K_{a} &=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{CHCl}_{2} \mathrm{COO}^{-}\right]}{\left[\mathrm{CHCl}_{2} \mathrm{COOH}\right]} \\ &=\frac{0.1 \times 0.02}{(0.08-0.02)}=3.3 \times 10^{-2} \end{aligned}

Posted by

Rakesh

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