Tag: nature of things

Questions Related to nature of things

The solubility of $A {2}X _{3}$ is $y$ $mol$ $dm^ {-3}$. Its solubility product is ____________.

chemistry nature of things objects that float or sink substances that sink or float soluble and insoluble substances

  1. $6y^ {4}$

  2. $64y^ {4}$

  3. $36y^ {5}$

  4. $108y^ {5}$

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Correct Option: A

What is the solubility of $Al(OH _{3}), K _{sp} = 1 \times 10^{-33}$, in a solution having $pH = 4$?

chemistry nature of things objects that float or sink substances that sink or float soluble and insoluble substances

  1. $6 \times 10^{-3}$M

  2. $10^{-6}$M

  3. $1.5 \times 10^{-4}$M

  4. $2.47 \times 10^{-9}$M

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Correct Option: D
Explanation:

The solubility equilibrium of $Al(OH) _3$ is

$Al(OH) _3$ ${\rightarrow} _{\leftarrow}$ $Al^{+3}$  + $3OH^-$
Therefore $x=1,\ y=2$ and $K _{sp}$=$[Al^{+3}]\times [OH^-]$ = $27S^4$
$S$ = $(\dfrac{10^{-33}}{27})^{1/4}$
Therefore $S$= $2.47\times 10^{-9}M$ in $ mol\ dm^{-3}$

Which of the following is most soluble?

chemistry nature of things objects that float or sink substances that sink or float soluble and insoluble substances

  1. ${Bi} _{2}{S} _{3} \left({K} _{sp} = 1\times {10}^{-17}\right)$

  2. $MnS\left({K} _{sp} = 7\times {10}^{-16}\right)$

  3. $CuS\left({K} _{sp} = 8\times {10}^{-37}\right)$

  4. ${Ag} _{2}S\left({K} _{sp} = 6\times {10}^{-51}\right)$

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Correct Option: A
Explanation:

Ksp (Solubility product constant) is the equilibrium between a solid and its respective ions in a solution. The value of the constant identifies the degree to which the compound can dissociate in water. The higher the Ksp the more soluble the compound is.
Bismuth is a large ion and sulphur is a small ion, the attractive forces between them are high thus the solubility is high. The solubility product can be calculated as :
$Bi _2S _3$
$K _sp = 1 \times 10^{-17}$
$K _sp = [Bi]^2 [S]^3$
$1 \times 10^{-17} = (2x)^2 (3x)^3 = 108x^5$
$x^5 = \dfrac{1 \times 10^{17}}{108} = 0.000156M $

The solubility of $Ag _{2}CO _{3}$ in water is $1.26 \times 10^{-4}$ mole/litre. What is solubility in 0.02 M $Na _{2}CO _{3}$ solution?
Note : Assume no hydrolysis of $CO _{3}^{2-}$ ion. (Take $\sqrt[3]{2} = 1.26$).

chemistry nature of things objects that float or sink substances that sink or float soluble and insoluble substances

  1. $5 \times 10^{-6}$M

  2. $\sqrt{50} \times 10^{-6}$M

  3. $10^{-5}$M

  4. $2 \times 10^{-5}$M

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Correct Option: C

$PbCl _2$ has maximum concentration of $1.0 \times 10^{-3}$ M in its saturated aq. solution at $25^0$C. Its solubility in $0.1$M NaCl solution will be : 

chemistry nature of things objects that float or sink substances that sink or float soluble and insoluble substances

  1. $4 \times 10^{-7}$M

  2. $4 \times 10^{-9}$M

  3. $2 \times 10^{-7}$M

  4. $2 \times 10^{-9}$M

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Correct Option: A
Explanation:

In saturated aqueous solution :
Maximum concentration in saturated aqueous solution is the solubility (S) of $\displaystyle PbCl _2$

$\displaystyle S=1.0 \times 10^{-3} \ M$

$\displaystyle [PbCl _2]=[Pb^{2+}]=S=1.0 \times 10^{-3} \ M$

$\displaystyle [Cl^-]=2 \times [PbCl _2]= 2 \times  S= 2 \times  1.0 \times 10^{-3} \ M=2.0 \times 10^{-3} \ M$

$\displaystyle K _{sp}=[Pb^{2+}] \times [Cl^-]^2$

$\displaystyle K _{sp}=1 .0 \times 10^{-3} \times [2.0 \times 10^{-3}]^2$

$\displaystyle K _{sp}=4.0 \times 10^{-9}$

In 0.1 M NaCl solution:

$\displaystyle [Cl^-]=[NaCl]=0.1 \ M$

Assuming $Cl^-$ ion from $PbCl _2$ is negligible

$\displaystyle K _{sp}=[Pb^{2+}] \times [Cl^-]^2$

$\displaystyle 4.0 \times 10^{-9}=[Pb^{2+}] \times [0.1]^2$

$\displaystyle [Pb^{2+}]=4.0 \times 10^{-7}  \ M$