How do you calculate Ka?

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How do you calculate Ka?

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Answer 1 · 2017-07-17T06:06:22

In fact you measure $K_{a}$ $K_a$ ........

Explanation:

For the hypothetical reaction..........

$A + B ⇌ C + D$ $A+BrightleftharpoonsC+D$

There is a $rate forward,$ $"rate forward,"$ $= k_{f} [A] [B]$ $=k_f[A][B]$ ......

And a $rate backward,$ $"rate backward,"$ $= k_{r} [C] [D]$ $=k_r[C][D]$ .......

And importantly the condition of chemical equilibrium does not specify the cessation of chemical change, but equality of forward and reverse rates (says he, channelling a text book)........

And thus $k_{f} [A] [B] = k_{r} [C] [D]$ $k_f[A][B]=k_r[C][D]$ ......

And $\frac{k_{f}}{k_{r}} = \frac{[C] [D]}{[A] [B]} = K_{eq}$ $k_f/k_r=([C][D])/([A][B])=K_"eq"$ , the thermodynamic equilibrium constant.

For acid base behaviour in water, we write......

$H A + H_{2} O ⇌ H_{3} O^{+} + A^{-}$ $HA+H_2OrightleftharpoonsH_3O^+ +A^-$

$K_{a} = \frac{[H_{3} O^{+}] [A^{-}]}{[H A]}$ $K_a=([H_3O^+][A^-])/([HA])$ , ............

the $[H_{2} O]$ $[H_2O]$ term may be removed, because it is so large it is effectively constant. Apologies if I have missed the point of your question.......but for strong acids,......

$H A = H X (X \neq F), H_{2} S O_{4}, H C l O_{4}$ $HA=HX (X!=F), H_2SO_4, HClO_4$ ,

.......the equilibrium lies to the right......And the extent of equilibrium, as always, is established by measurement.

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How do you calculate Ka?

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