# What is the pH of a solution prepared by mixing 50.0 mL of 0.010 mol/L HCl(aq) and 50.0 mL of 0.010 mol/L Ca(OH)2(aq)? Assume the temperature is 25 degrees celcius.

Apr 26, 2017

$p H = 11.4 \ldots \ldots . .$

#### Explanation:

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

$\text{Moles of HCl} = 50.0 \times {10}^{-} 3 L \times 0.010 \cdot m o l \cdot {L}^{-} 1 = 5.00 \times {10}^{-} 4 \cdot m o l$.

$\text{Moles of } C a {\left(O H\right)}_{2} = 50.0 \times {10}^{-} 3 L \times 0.010 \cdot m o l \cdot {L}^{-} 1 = 5.00 \times {10}^{-} 4 \cdot m o l$.

Now there are EQUIMOLAR quantities of $C a {\left(O H\right)}_{2}$ and $H C l$.

Clearly, at the end of the addition, we have a$100 \cdot m L$ solution that contains, nominally, $C a \left(O H\right) C l$, i.e. we have added HALF an EQUIV of acid to the base.

And thus we have $2.500 \times {10}^{-} 4 \cdot m o l$ $H {O}^{-}$ in $100 \cdot m L$ of solution.

$\left[H {O}^{-}\right] = 2.50 \times {10}^{-} 4 \cdot m o l \cdot {L}^{-} 1$.

$p O H = - {\log}_{10} \left(\left[H {O}^{-}\right]\right) = - {\log}_{10} \left(2.50 \times {10}^{-} 4\right) = - \left(- 2.60\right) = 2.60$.

But $p H = 14 - p O H = 14 - 2.60 = 11.4$.

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