# What is the pH of a 0.026 M Sr(OH)2 solution?

May 27, 2018

Well, what's the $\text{pOH}$? I get $\text{pOH} = 1.28$. What then is the $\text{pH}$ at ${25}^{\circ} \text{C}$?

We assume "Sr"("OH")_2 is a strong base, such that

${\text{Sr"("OH")_2(s) stackrel("H"_2"O"(l)" ")(->) "Sr"^(2+)(aq) + 2"OH}}^{-} \left(a q\right)$

and thus, it supposedly gives rise to $0.026 \times 2 = {\text{0.052 M OH}}^{-}$. As a result,

"pOH" = -log["OH"^(-)] = -log(0.052) = 1.28

But clearly, we have the $\text{pOH}$ and not the $\text{pH}$. At any temperature,

${\text{pH" + "pOH" = "pK}}_{w}$,

and at ${25}^{\circ} \text{C}$, ${\text{pK}}_{w} = 14$. Therefore:

$\textcolor{b l u e}{\text{pH}} = 14 - 1.28 = \textcolor{b l u e}{12.72}$

But as chemists, we must check the data... The ${K}_{s p}$ of "Sr"("OH")_2 is around $6.4 \times {10}^{- 3}$. The ICE table gives:

${\text{Sr"("OH")_2(s) rightleftharpoons "Sr"^(2+)(aq) + 2"OH}}^{-} \left(a q\right)$

$\text{I"" "-" "" "" "" "" "0" "" "" "" } 0$
$\text{C"" "-" "" "" "" "+s" "" "" } + 2 s$
$\text{E"" "-" "" "" "" "" "s" "" "" "" } 2 s$

This is equal to the mass action expression:

${K}_{s p} = 6.4 \times {10}^{- 3} = {\left[{\text{Sr"^(2+)]["OH}}^{-}\right]}^{2}$

$= s {\left(2 s\right)}^{2} = 4 {s}^{3}$

$= \frac{1}{2} {\left(2 s\right)}^{3} = \frac{1}{2} {\left[{\text{OH}}^{-}\right]}^{3}$

And so, the maximum concentration of ${\text{OH}}^{-}$ at ${25}^{\circ} \text{C}$ is:

$\left[{\text{OH}}^{-}\right] = {\left(2 {K}_{s p}\right)}^{1 / 3}$

$= {\left(2 \cdot 6.4 \times {10}^{- 3}\right)}^{1 / 3}$

$=$ $\text{0.234 M}$

And since $\text{0.052 M}$ $<$ $\text{0.234 M}$, the concentration given in the question is valid and physically realizable.