Question 23670

Apr 6, 2015

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

$H C {l}_{\left(a q\right)} \to {H}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

As you can see, 1 mole of $H C l$ produces 1 mole of ${H}^{+}$ and 1 mole of $C {l}^{-}$ ions.

Use the molarity of the two solutions to determine how many moles of each you mix

$C = \frac{n}{V} \implies n = C \cdot V$

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 $C {l}^{-}$ ions to the mix.

The total volume of the new solution will be

${V}_{\text{total") = V_("sol 1") + V_("sol 2}}$

V_("total") = "16.0" + 12.0 = "28.0 cm"^(3)

The total number of ${H}^{+}$ and $C {l}^{-}$ moles will be

${n}_{{H}^{+}} = {n}_{\text{sol 1") + n_("sol 2}}$
${n}_{{H}^{+}} = 0.00224 + 0.009 = {\text{0.01124 moles H}}^{+}$

and

${n}_{C {l}^{-}} = {n}_{\text{sol 1") + n_("sol 2}}$
${n}_{C {l}^{-}} = 0.00224 + 0.009 = {\text{0.01124 moles Cl}}^{-}$

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")#