A solution is prepared to be 0.1475 M in Sr(OH)2. How do you find the [H3O+], [OH‐], pH, and pOH?

Jun 10, 2016

$p H = 13.029 \text{ and } p O H = 0.9710$

Explanation:

Strontium Hydroxide ionizes in water to produce the strontium ion and the hydroxide ion. Note that $S r {\left(O H\right)}_{2}$ is slightly soluble in water.

$\text{ } S r {\left(O H\right)}_{2} \setminus \setminus r i g h t \le f t h a r p \infty n s \setminus \setminus \setminus S {r}^{2 +} + \setminus \setminus \setminus 2 O {H}^{-}$

$I \text{ " 0.1475 \ M " " - " } -$

$C \text{ " -x " "+x " } + 2 x$

$E \text{ "0.1475-x " " x " } 2 x$

K_b=( [ OH^-]^2 [ Sr^(2+)])/([Sr(OH)_2]

${K}_{b} = \frac{{\left(2 x\right)}^{2} \cdot \left(x\right)}{0.1475 - x} = 6.5 \times {10}^{-} 3$

Solve for $x$

$x = 0.05346 \setminus M$

$\left[O {H}^{-}\right] = 2 x = 2 \times 0.05346 = 0.1069 M$

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In any aqueous solution, both ions ${H}_{3} {O}^{+}$and $O {H}^{-}$ are present and they must satisfy the following condition:

$\left[{H}_{3} {O}^{+}\right] \left[O {H}^{-}\right] = {K}_{W}$

$\left[{H}_{3} {O}^{+}\right] \left[O {H}^{-}\right] = 1.0 \times {10}^{- 14}$

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$\left[O {H}^{-}\right] = 0.1069 \setminus M$ ( determined)

$\left[{H}_{3} {O}^{+}\right] = \frac{1.0 \times {10}^{-} 14}{\left[O {H}^{-}\right]}$

$\left[{H}_{3} {O}^{+}\right] = \frac{1.0 \times {10}^{-} 14}{0.1069}$

$\left[{H}_{3} {O}^{+}\right] = {9.355}^{-} 14 \setminus M$

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$p H = - \log \left[{H}_{3} {O}^{+}\right]$

$p H = - \log \left[\setminus {3.955}^{-} 14 \setminus\right]$

$p H = 13.0290$

$p O H = - \log \left[\setminus O {H}^{-} \setminus\right]$

$p O H = - \log \left[\setminus 0.1069 \setminus\right]$

$p O H = 0.9710$