Question

A 0.10 M HF solution is 8.4% ionized. Calculate the H+ ion concentration

Answer 1

0.10 M HF solution is 0.10 mol HF / L H₂O

Let's look at one liter of solution. That liter has 0.10 mol HF. 8.4% of this is ionized. Therefore, 0.0084 mol of the total areionized into H⁺and F^-. Since this is in one liter, theH⁺concentration is0.0084 mol/L =0.0084 M

Answer 2

Let α be the dissociation of the weak acid ,HF
                            HF <---> H + + F-

initial conc.            c               0         0

Equb. conc.         c(1-α)        cα      cα

Dissociation constant , Ka = cα * cα / ( c(1-α)

                                         = c α^2 / (1-α)

In the case of weak acids α is very small so 1-α is taken as 1

So Ka = cα^2

 α = √ ( Ka / c )

Given c = concentration = 0.10 M

% dissociation =  α*100 = 8.4 

 α = 8.4*10^-2

 [H+] = cα = 0.10 * 8.4*10^-2 = 8.4*10^-3 M