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Let be monoprotic acid be HA. It will dissociate as follows :
HA -------> H+ + A-
1 moles of H+ is formed from 1 mole of acid
Molar concentration of the acid = Molarity = moles/ L of solution
number of moles of acid present = 0.003g/ 128 g mol-1 = 2.34 *10-5
pH = - log [H+] = 4.98
[H+] = 1.05 *10-5 M
[H+] = [A-] = 1.05 *10-5
Keq = [H+][A-]/[HA]
= (1.05 *10-5)2 / 2.34 *10-5
= 0.47 *10-5
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