Question #23670

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Question #23670

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Answer 1 · 2015-04-06T09:56:09

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

$H C l_{(a q)} \to H_{(a q)}^{+} + C l_{(a q)}^{-}$ $HCl_((aq)) -> H_((aq))^(+) + Cl_((aq))^(-)$

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

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

$C = \frac{n}{V} \Rightarrow n = C \cdot V$ $C = n/V => n = C * V$

$n_{sol 1} = 0.140 M \cdot 16.0 \cdot 10^{- 3} L = 0.00224 moles HCl$ $n_("sol 1") = "0.140 M" * 16.0 * 10^(-3)"L" = "0.00224 moles HCl"$

and

$n_{sol 2} = 0.750 M \cdot 12.0 \cdot 10^{- 3} L = 0.009 moles HCl$ $n_("sol 2") = "0.750 M" * 12.0 * 10^(-3)"L" = "0.009 moles HCl"$

Automatically, each solutions brings that many $H^{+}$ $H^(+)$ and $C l^{-}$ $Cl^(-)$ ions to the mix.

The total volume of the new solution will be

$V_{total} = V_{sol 1} + V_{sol 2}$ $V_("total") = V_("sol 1") + V_("sol 2")$

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

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

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

and

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

The new concentrations will be

$C_{H^{+}}^{+} = \frac{0.01124 moles}{28.0 \cdot 10^{- 3} L} = 0.401 M$ $C_(H^(+)) = "0.01124 moles"/(28.0 * 10^(-3)"L") = color(green)("0.401 M")$

$C_{C l^{-}}^{-} = \frac{0.01124 moles}{28.0 \cdot 10^{- 3} L} = 0.401 M$ $C_(Cl^(-)) = "0.01124 moles"/(28.0 * 10^(-3)"L") = color(green)("0.401 M")$

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Question #23670

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