# Question 68f3f

Apr 17, 2015

One must first work out the pOH which can be determined from the concentration of the OH ion.

To do this you first have to calculate the concentration of the Ca(OH)2 and KOH in the new volume which would be 25ml (10ml +15ml). So C1V1 = C2V2 e.g 2.4*10^-6 * 10ml = C2 * 25ml. Then it's easy to work out C2.

After you have done that for both compounds times the Ca(OH)2 concentration by 2 because there are two moles of OH. Add the concentration of KOH which would be the concentration of the OH to the answer worked out before (OH concentration) and that would be the final concentration of OH.

Finally work out the pOH by using pOH= - Log [OH] then take 14 - pOH to obtain the pH, which is your answer.

Apr 17, 2015

The pH of the solution will be 11.08.

So, you're essentially mixing two solutions that contain strong bases, calcium hydroxide, $C a {\left(O H\right)}_{2}$, and potassium hydroxide, $K O H$.

SIDE NOTE SInce you didn't specify what $2.4 \cdot {10}^{- 6}$ and $2.5 \cdot {10}^{- 5}$ mean, I'll assume that they are the number of moles of each substance present in that solution.

Although calcium hydroxide is not very soluble in water, it is still considered a strong base because what does dissolve will dissocite completely to give

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

Notice that you get $\textcolor{red}{2}$ moles of hydroxide ions for every mole of calcium hydroxide that dissociates.

You need to keep track of the number of moles of hydroxide ions that you add together. Potassium hydroxide dissociates completely in aqueous solution to give

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

This time, 1 mole of potassium hydroxide produces 1 mole of hydroxide ions.

So, you've got 1 mole of hydroxide ions for every mole of $K O H$ and 2 moles of hydroxide ions for every mole of $C a {\left(O H\right)}_{2}$, which means that the total number of moles of hydroxide will be

${n}_{\text{total} O {H}^{-}} = {n}_{K O H} + 2 \cdot {n}_{C a {\left(O H\right)}_{2}}$

n_("total" OH^(-)) = 2.5 * 10^(-5) + 2 * 2.4 * 10^(-6) = 2.98 * 10^(-5)"moles OH"^(-)

Don't forget that the total volume of the final solution will be

${V}_{\text{final}} = {V}_{K O H} + {V}_{C a {\left(O H\right)}_{2}}$

${V}_{\text{final" = 10 + 15 = "25 mL}}$

You know how many moles of $O {H}^{-}$ ions you have, and the volume of the solution, so you can calculate the concentration of the ions

C = n/V = (2.98 * 10^(-5)"moles")/(25 * 10^(-3)"L") = 1.19 * 10^(-3)"M"#

Solve for pOH by using

$p O H = - \log \left(\left[O {H}^{-}\right]\right) = - \log \left(1.19 \cdot {10}^{- 3}\right) = 2.92$

The pH of the solution will be

$p {H}_{\text{sol}} = 14 - p O H = 14 - 2.92 = \textcolor{g r e e n}{11.08}$