Calculate the pH and the concentrations of all species present, #"H"_3"O"^(+)#, #"F"^(-)#, #"HF"#, and #"OH"^(-)#, in #"0.05 M HF"# ?

For starters, make sure that you have the value of the acid dissociation constant for hydrofluoric acid

#K_a = 7.2 * 10^(-4)#

http://clas.sa.ucsb.edu/staff/Resource%20Folder/Chem109ABC/Acid,%20Base%20Strength/Table%20of%20Acids%20w%20Kas%20and%20pKas.pdf

So, hydrofluoric acid is a weak acid that only partially ionizes in aqueous solution to produce hydronium cations and fluoride anions

#"HF"_ ((aq)) + "H"_ 2"O" _ ((l)) rightleftharpoons "H"_ 3 "O"_ ((aq))^(+) + "F"_ ((aq))^(-)#

Notice that the ionization equilibrium produces #1# mole of hydronium cations and #1# mole of fluoride anions for every #1# mole of hydrofluoric acid that ionizes.

If you take #x# to be the equilibrium concentration of the hydronium cations, you can say that

#["H"_3"O"^(+)] = ["F"^(-) ] =x# #"M " -># the two ions are produced in a #1:1# mole ratio

#["HF"] = ["HF"]_0 - x -># the concentration of the acid decreases by #x# #"M"#

In your case

#["HF"]_0 = "0.05 M"#

and so the equilibrium concentration of the acid, i.e. what remains unionized in solution, will be

#["HF"] = (0.05 - x)# #"M"#

By definition, the acid dissociation constant is equal to

#K_a = (["H"_3"O"^(+)] * ["F"^(-)])/(["HF"])#

In your case, you have

#7.2 * 10^(-4) = (x * x)/(0.05 - x) = x^2/(0.05 - x)#

Rearrange to quadratic equation form

#x^2 + 7.2 * 10^(-4) * x - 0.05 * 7.2 * 10^(-4) = 0#

#color(white)(a)/color(white)(aaaaaaaaaaaaaaaa)#

SIDE NOTE: You cannot use the approximation

#0.05 - x ~~ 0.05#

because you will have

#7.2 * 10^(-4) = x^2/0.05#

which will get you

#x = 0.006#

Remember that the approximation holds only if

#x/(["HF"]_0) xx 100% < 5%#

In this case, the approximation does not hold, since

#(0.006 color(red)(cancel(color(black)("M"))))/(0.005color(red)(cancel(color(black)("M")))) xx 100% = 12% > 5%#

#color(white)(a)/color(white)(aaaaaaaaaaaaaaaa)#

This quadratic equation will produce two solutions, one positive and one negative. Since #x# represents concentration, the negative solution does not have a physical significance in this context.

Therefore, you will have

#x = 0.00656#

This means that at equilibrium, you will have

#["H"_ 3"O"^(+)] = "0.00656 M"#

#["F"^(-)] = "0.00656 M"#

#["HF"] = "0.05 M" - "0.00565 M" = "0.0444 M"#

As you know, an aqueous solution at room temperature has

#color(blue)(ul(color(black)(["H"_ 3"O"^(+)] * ["OH"^(-)] = 10^(-14))))#

This means that you will have

#["OH"^(-)] = 10^(-14)/(["H"_3"O"^(+)])#

which gets you

#["OH"^(-)] = 10^(-14)/(0.00565) = 1.77 * 10^(-12)# #"M"#

Finally, the pH of the solution

#color(blue)(ul(color(black)("pH" = - log( ["H"_3"O"^(+)]))))#

will be equal to

#"pH" = - log(0.00565) = 2.25#

I'll leave the concentrations rounded to three sig figs and the pH rounded to two decimal places, but keep in mind that you only have one significant figure for the initial concentration of the acid.