I will assume a value (i), and I happen to know (ii).

We assume that we have a solution that is FORMALLY #1*mol*L^-1# with respect to acetic acid.......(this is approx. the same concentration as household vinegar, the which of course is aqueous acetic acid).

In aqueous solution, acetic acid undergoes some degree of dissociation:

#HOAc(aq) +H_2O(l)rightleftharpoonsH_3O^+ + ""^(-)OAc#

Now the equilibrium is governed by an equilibrium constant, #K_a#, which must be known or measured; for this problem it should have been quoted, #K_a=10^(-4.76)#

#K_a=10^(-4.76)=([H_3O^+][""^(-)OAc])/([HOAc])#

We ASSUME that a concentration of #x_1# moles of acetic acid dissociates........and so....

#K_a=10^(-4.76)=(x_1xxx_1)/(1-x_1)#

This is a quadratic in #x_1#, which we could solve exactly with the quadratic equation. Because we are lazy, we ASSUME that #1-x_1~=1#

And thus #x_1=sqrt(10^(-4.76))=4.17xx10^-3*mol*L^-1#.

#x_1# is indeed small compared to #1#, but we can recycle the expression, and use this first approximation.....

#K_a=10^(-4.76)=(x^2)/(1-4.17xx10^-3)#

#x_2=4.17xx10^-3*mol*L^-1#.

And given #x=[H_3O^+]#, and............... #pH=-log_10[H_3O^+]=-log_10(4.17xx10^-3)=2.38#

Easy, ain't it? But you can get very good and very systematic at these sorts of problems......