# Question 11a0b

Apr 15, 2017

$\left[O {H}^{-}\right] = 0.125 M$
$\left[{H}^{+}\right] = 8 \times {10}^{- 14} M$
and $p H = 13.1$

#### Explanation:

Both $H C l$ and $C a {\left(O H\right)}_{2}$ are strong electrolytes, so each will ionize 100% in solution.

This makes it possible to determine the amount of ${H}^{+}$ in the $H C l$ solution and the amount of $O {H}^{-}$ in the $C a {\left(O H\right)}_{2}$ solution, by multiplying concentration x volume in each case:

$H C l$: (0.250 "mol"/L)(0.500 L) = 0.125 "mol" of HCl

and since it is strong, moles ${H}^{+} = 0.125$ as well.

$C a {\left(O H\right)}_{2}$: (0.250 "mol"/L)(0.500 L) = 0.125 "mol" 

and since it is strong and because each one mole of $C a {\left(O H\right)}_{2}$ contains two moles of $O {H}^{-}$ ions, moles $O {H}^{\equiv} 0.250$.

Looking at neutralization, we see that the reaction is one-to-one between ${H}^{+}$ and $O {H}^{-}$

${H}^{+} + O {H}^{-}$$\rightarrow {H}_{2} O$

So, the 0.125 mol of ${H}^{+}$ consumes an equal amount of $O {H}^{-}$, but 0.125 mol $O {H}^{-}$ will remain, now in a total volume of 1.0 L

Therefore

$\left[O {H}^{-}\right] = 0.125 \frac{\text{mol}}{1.0} L = 0.125 M$

Since [OH^-] = 0.125 M,

$\left[{H}^{+}\right] = \frac{1 \times {10}^{- 14}}{0.125} = 8 \times {10}^{- 14} M$

and the pH is -log(8 xx 10^(-14)) = 13.1