0.195 mol sample of HX is dissolved in water to form 645.0 mL solution
concentration of the solution = 0.195/0.645 = 0.30 M
now pH of the solution = -log[H3O+] = 4.30
so [H3O+] = 10^-4.3 = 5.01*10^-5 M
now, [H3O+]^2 =Ka* concentration of the solution
Ka = 8.37*10^-9
A 0.195-mol sample of HX is dissolved in enough H2O to form 645.0 mL of solution....
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