# Question #84342

Dec 16, 2014

SInce $H C l$ is a strong acid and $B a {\left(O H\right)}_{2}$ is a strong base, you are dealing with a neutralization reaction.

Let's start by writing the balanced chemical equation

$2 H C {l}_{\left(a q\right)} + B a {\left(O H\right)}_{2 \left(a q\right)} \to B a C {l}_{2} \left(a q\right) + 2 {H}_{2} {O}_{\left(l\right)}$

Strong acids and strong bases dissociate completely in aqueous solution, so the reaction's complete ionic equation is

$2 {H}_{\left(a q\right)}^{+} + 2 C {l}_{\left(a q\right)}^{-} + B {a}_{\left(a q\right)}^{2 +} + 2 O {H}_{\left(a q\right)}^{-} \to B {a}_{\left(a q\right)}^{2 +} + 2 C {l}_{\left(a q\right)}^{-} + 2 {H}_{2} {O}_{\left(l\right)}$

After eliminating the spectator ions - the ions that can be found both on the reactants, and on the products' side - the net ionic equation is

${H}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-} \to {H}_{2} {O}_{\left(l\right)}$

Let's determine the number of moles for each reactant using their respective concentration, $C = \frac{n}{V}$

${n}_{H C l} = {C}_{H C l} \cdot {V}_{H C l} = 0.10 M \cdot 50.0 \cdot {10}^{- 3} L = 5.0 \cdot {10}^{- 3}$ moles

${n}_{B a {\left(O H\right)}_{2}} = C \cdot V = 0.20 M \cdot 150.0 \cdot {10}^{- 3} L = 30 \cdot {10}^{- 3}$ moles

We've determined from the balanced chemical equation that 2 moles of $H C l$ need 1 mole of $B a {\left(O H\right)}_{2}$; however, notice that we have more moles of $B a {\left(O H\right)}_{2}$ than of $H C l$, which means that the mixture will contain an excess of $B {a}^{2 +}$ and $O {H}^{-}$ ions.

The final concentrations will be

${C}_{H C l} = \frac{n}{V} _ \left(T O T A L\right) = \frac{5.0 \cdot {10}^{- 3} m o l e s}{\left(50.0 + 150.0\right) \cdot {10}^{- 3} L} = 0.025 M$

${C}_{B a {\left(O H\right)}_{2}} = \frac{30.0 \cdot {10}^{- 3} m o l e s}{\left(50.0 + 150.0\right) \cdot {10}^{- 3} L} = 0.15 M$

Therefore,

$\left[{H}^{+}\right] = 0.025 M$
$\left[C {l}^{-}\right] = 0.025 M$
$\left[B {a}^{2 +}\right] = 0.15 M$
$\left[O {H}^{-}\right] = 0.15 M$