# Question #3304e

##### 1 Answer
Dec 13, 2014

The answer is pH = $11.85$.

The balanced chemical equation is

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

SInce $N a O H$ is a strong base and $H C l$ is a strong acid, they will dissociate completely in aqueous solution; likewise, $N a C l$, being a salt, will dissociate completely in $N {a}^{+}$ and $C {l}^{-}$ ions, thus giving us the complete ionic equation

$N {a}_{\left(a q\right)}^{+} + O {H}_{\left(a q\right)}^{-} + C {l}_{\left(a q\right)}^{-} + {H}_{3}^{+} {O}_{\left(a q\right)} \to 2 {H}_{2} {O}_{\left(a q\right)} + N {a}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

SInce $N {a}^{+}$ and $C {l}^{-}$ ions are present both on the reactants and on the products' side, they are considered to be spectator ions - which leads us to the net ionic equation

${H}_{3}^{+} {O}_{\left(a q\right)} + O {H}_{\left(a q\right)}^{-} \to 2 {H}_{2} {O}_{\left(l\right)}$ - this represents the neutralization reacton between $N a O H$ and $H C l$.

You could consider this to be a titration problem - let's assume we are titrating $120.0 m L$ of $H C L$, $0.0150 M$, with $70.0 m L$ of $N a O H$, $0.0450 M$.

We know the volume of $H C l$ to be 120.0mL, which translates to a number of moles of

${n}_{H C l} = C \cdot V = 0.0150 M \cdot 120.0 \cdot {10}^{-} 3 L = 0.0018$ moles

Now we add the 70.0mL of $N a O H$. The number of $N a O H$ moles added is

${n}_{N a O H} = C \cdot V = 0.0450 M \cdot 70.0 \cdot {10}^{-} 3 L = 0.00315$ moles

Comparing the number of $H C l$ and $N a O H$ moles, which equal the number of ${H}_{3}^{+} O$ and $O {H}^{-}$ moles, we can see that we have an excess of $O {H}^{-}$ moles of

${n}_{e x c e s s O {H}^{-}} = {n}_{O {H}^{-}} - {n}_{{H}_{3}^{+} O} = 0.00315 - 0.0018 = 0.00135$ moles left in the solution. This gives us the concentration of $O {H}^{-}$

$\left[O {H}^{-}\right] = {n}_{e x c e s s} / \left({V}_{T O T A L}\right) = \frac{0.00135 m o l e s}{\left(120.0 + 70.0\right) \cdot {10}^{-} 3 L} = 0.0071 M$

(${V}_{T O T A L}$ represents the combined volume of the new solution)

So, $p O H = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left(0.0071\right) = 2.15$

Therefore, pH $= 14 - p O H = 14 - 2.15 = 11.85$