No, #"HQuCl"# is not the same as #"HQu"^(+)#, since the former is a salt and the latter is the conjugate acid of your weak base, #"Qu"#.
#"HQuCl"# however will dissociate in aqueous solution to give #"HQu"^(+)# and #"Cl"^(-)#.
#HQuCl_((aq)) rightleftharpoons HQu_((aq))^(+) + Cl_((aq))^(""-)#
The reaction of interest will be the dissociation of the conjugate acid to form your original base and the hydronium ion
#HQu_((aq))^(+) + H_2O_((l)) rightleftharpoons Qu_((aq)) + H_3O_((aq))^(+)#
This is the reaction you'll use to calculate the pH of the solution. Do not forget to use #pK_a#, not #pK_b#, for this equilibrium. Remember that
#pK_a + pK_b = pK_w#, or
#pK_a + pK_b = 14 => pK_a = 14 - pK_b#
Use this value to determine the value of the acid dissociation constant, #K_a#, by
#K_a = 10^(-pKa)#
Now you've got all you need to calculate the pH of the solution.
