Tag: soluble and insoluble substances

Questions Related to soluble and insoluble substances

At ${25}^{o}C$ the solubility of ${Ag} _{2}{CO} _{3}$ (${K} _{sp}=4.3\times {10}^{-13}$) would be in what order in the following solutions?

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

  1. $0.01M$ $Ag{NO} _{3}$

  2. $0.04M$ ${K} _{2}{CO} _{3}$

  3. pure water

  4. in a buffer ($pH=4$)

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Correct Option: A

$Ca _3(PO _4) _2$ is insoluble in water. On adding a few drops of HCl to solid $Ca _3(PO _4) _2$ in contact with water, the solid dissolves. The reason is?

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

  1. The solvent becomes more polar on adding HCl

  2. $Ca _3(PO _4) _2$ combines with HCl to form soluble $CaCl _2$

  3. $Ca(H _2PO _4) _2$ is formed, which dissolves

  4. $H _3PO _4$, a weak acid is formed and the solubility product of $Ca _3(PO _4) _2$ decrease

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Correct Option: A

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

Show answer
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

Show answer
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

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

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

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