This time, you add two solutions together and want the concentration of each ion present.
Because hydrochloric acid is a strong acid, it dissociates completely in aqueous solution to give
#HCl_((aq)) -> H_((aq))^(+) + Cl_((aq))^(-)#
As you can see, 1 mole of #HCl# produces 1 mole of #H^(+)# and 1 mole of #Cl^(-)# ions.
Use the molarity of the two solutions to determine how many moles of each you mix
#C = n/V => n = C * V#
#n_("sol 1") = "0.140 M" * 16.0 * 10^(-3)"L" = "0.00224 moles HCl"#
and
#n_("sol 2") = "0.750 M" * 12.0 * 10^(-3)"L" = "0.009 moles HCl"#
Automatically, each solutions brings that many #H^(+)# and #Cl^(-)# ions to the mix.
The total volume of the new solution will be
#V_("total") = V_("sol 1") + V_("sol 2")#
#V_("total") = "16.0" + 12.0 = "28.0 cm"^(3)#
The total number of #H^(+)# and #Cl^(-)# moles will be
#n_(H^(+)) = n_("sol 1") + n_("sol 2")#
#n_(H^(+)) = 0.00224 + 0.009 = "0.01124 moles H"^(+)#
and
#n_(Cl^(-)) = n_("sol 1") + n_("sol 2")#
#n_(Cl^(-)) = 0.00224 + 0.009 = "0.01124 moles Cl"^(-)#
The new concentrations will be
#C_(H^(+)) = "0.01124 moles"/(28.0 * 10^(-3)"L") = color(green)("0.401 M")#
#C_(Cl^(-)) = "0.01124 moles"/(28.0 * 10^(-3)"L") = color(green)("0.401 M")#
