# Question f3d82

May 13, 2015

Keep in mind, however, that both compounds will contribute to the number of moles of hydroxide ions present in the final solution.

Since you're mixing two strong bases, which by definition dissociate completely in aqueous solution, you'll get

$K O H \to {K}^{+} + O {H}^{-}$

and

$C a {\left(O H\right)}_{2} \to C {a}^{2 +} + \textcolor{red}{2} O {H}^{-}$

Calcium hydroxide will produce 2 moles of hydroxide ions for every mole of $C a {\left(O H\right)}_{2}$.

This means that the total number of moles of $O {H}^{-}$ will be

${n}_{O {H}^{-}} = 0.572 \cdot {10}^{- 3} + \textcolor{red}{2} \cdot 0.91 \cdot {10}^{- 3}$

${n}_{O {H}^{-}} = \left(0.572 + 1.82\right) \cdot {10}^{- 3} = 2.39 \cdot {10}^{- 3} \text{moles}$ $O {H}^{-}$

This is the number of moles you'll use for the final concentration of hydroxide ions

$\left[O {H}^{-}\right] = \left(2.39 \cdot \cancel{{10}^{- 3}} \text{moles")/(15.2 * cancel(10^(-3))"L") = color(green)("0.157 M}\right)$

May 13, 2015

In order to avoid a lot of zeroes and ten-powers I usually work with milliMoles (=1/1000 Mole). You will see that in the end the milli's cancel out.
Other note: I use $m L$ in stead of $c {m}^{3}$ (they're the same volume).

First solution ($K O H$):
$5.20 m L \cdot 0.110 M = 0.572 m M o l$

Second solution ($C a {\left(O H\right)}_{2}$):
$10.0 m L \cdot 0.091 M = 0.91 m M o l$
Now remember $C a {\left(O H\right)}_{2}$ gives off two $O {H}^{-}$ on dissociation (that's where you went wrong), so the total

OH^-)=0.572+2*0.91=2.392mMol#

This is in a volume of $5.20 + 10.0 = 15.20 m L$

So $\left[O {H}^{-}\right] = \frac{2.392 m M o l}{15.20 m L} = 0.157 M o l / L = 0.157 M$