CH3COOH + H2O --> H3O+ + CH3COO-
initial 0.010 -- ? ?
Equilibrium 0.010-x -- ? ?
Show how did you solve this equation to get 1.76E-5:
Ka = [H3O][CH3COO]/[CH3COOH] = 1.76E-5
(b) 0.010 M hydrochloric acid
(c) Which acid solution in Question#1(a) or #1(b) has the lower pH ?
(d) Why does the acid solution noted in (c) have a lower pH ?
a) If the initial conc. of a weak acid is C moles/litre and if x is the degree of dissociation (defined as the fraction of the total electrolyte split up at equilibrium), then, the equilibrium concentrations would be,
<br>C (1-x) mole/litre, of the undissociated acid and Cx moles/litre each of H+ and A-
<br>the equilibrium constant for the ionization process is given ...
The expert calculates the pH.