# (a) The molar solubility of PbBr_{2}\ at\ 25^{\circ}C\ is\ 1.0 \tim

(a) The molar solubility of . Calculate Kp- (b) If 0.0490 g of $AgI{O}_{3}$ dis- sp• solves per liter of solution, calculate the solubility-product constant. (c) Using the appropriate Ksp value from Appen- dix D, calculate the pH of a saturated solution of $Ca{\left(OH\right)}^{2}$.

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Step 1
The molar solubility of . The amount of $AgI{O}_{3}$ dissolved per liter of solution is 0.0490 g. The value of .
Step 2
(a) The value of ${K}_{sp}$ for compound with general formula $A{X}_{2}$ is calculated as,
${K}_{sp}=4{s}^{3}$
Where,
- s is the molar solubility of compound.
Substitute the value of molar solubility of PbBr2 in the above formula.
${K}_{sp}=4{\left(1.0×{10}^{-2}\right)}^{3}$
$=4.0×{10}^{-6}$
The solubility product constant of .
Step 3
(b) The number of moles of $AgI{O}_{3}$ is calculated by the formula,
${n}_{AgI{O}_{3}}=\frac{m}{M}$
Where,
- m is the given mass of $AgI{O}_{3}$.
- M is the molar mass of $AgI{O}_{3}$.
The molar mass of .
Substitute the given mass and molecular mass of $AgI{O}_{3}$ in the above formula.
${n}_{AgI{O}_{3}}=\frac{0.0490g}{282.77\frac{g}{}mol}$
$=1.73×{10}^{-4}mol$
Hence, the molar solubility of .
The value of ${K}_{sp}$ for compound with general formula AX is calculated as,
${K}_{sp}={s}^{2}$
Substitute the molar solubility of $AgI{O}_{3}$ in the above formula.
${K}_{sp}={\left(1.73×{10}^{-4}\right)}^{2}$
$=3.0×{10}^{-8}$
The solubility product constant of .
Step 4
(c) The value of Ksp for compound with general formula $A{X}_{2}$ is calculated as,
${K}_{sp}=4{s}^{3}$
Substitute the value of in the above formula.
$6.5×{10}^{-6}=4{s}^{3}$

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Elois Puryear

(a) $\text{molar solubility}=1.0×{10}^{-2}\frac{mol}{L}$
$PbB{r}_{2}⇒P{b}_{2}+2B{r}^{-}$
$\left[P{b}^{2+}\right]=1.0×{10}^{-2}\frac{mol}{L}$
$\left[Br-\right]=2×1.0×{10}^{-2}\frac{mol}{L}=2.0×{10}^{-2}\frac{mol}{L}$
${K}_{sp}=\left[P{b}^{2+}\right]{\left[Br-\right]}^{2}=\left(1.0×{10}^{-2}\right){\left(2.0×{10}^{-2}\right)}^{2}=4.0×{10}^{-6}$
enable the molar solubilit be S Then $\left[C{a}^{2+}\right]=S{\left[I{O}_{3}\right]}^{2-}=2S$ The solubility product which on fixing yields $S=8.9×{10}^{-3}$ and the molar solubility For the subsequent party, you consider x because the molar solubility of $Ca{\left(I{O}_{3}\right)}^{2}$ in water and then the most objective of IO3 must be $\left(2x+0.06\right)$ for the reason that $NaI{O}_{3}$ is carefully soluble and then $\left[C{a}^{2+}\right]=x$ then writing the solubility product equation and be certain for x, it quite is the molar solubility $Ca{\left(I{O}_{3}\right)}^{2}$ in the presence of .
(b) The reaction $AgI{O}_{3}\left(s\right)⇌A{g}^{+}\left(aq\right)+I{O}_{3}^{-}$
0.0490g of $AgI{O}_{3}$ dissolves per liter of solution.
Let us calculate the solubility-product constant.
First, let us find the molar solubility of $AgI{O}_{3}$.
The molar mass of $AgI{O}_{3}$ is
${M}_{AgI{O}_{3}}={M}_{Ag}+{M}_{I}+3\cdot {M}_{O}$