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The pH of 0.1 M monobasic acid is 4.50. Calculate the concentration of species H+ , A and HA at equilibrium. Also, determine the value of Ka and pKa of the monobasic acid.

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Given 

C = 0.1 M

pH = 4.50

\(\therefore\) pH = – log [H+]

or [H+] = 10–pH = 10–4.50

= 3.16 × 10–5

[H+] = [A] = 3.16 × 10–5

For the reaction,

HA ⇌ H+ + A

Ka\(\frac{[H^+][A^-]}{[HA]}\)

\(\frac{(3.16\times10^{-15})(3.16\times10^{-5})}{0.1}\)

[∵ [HA]equi = 0.1 – (3.16 × 10–5 ) ~ 0.1]

= 1.0 × 10–8

\(\because\) pKa = – log [Ka]

= – log (1 × 10–8 )

= 8 log 10

= 8

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