Tag: objects that float or sink

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Multiple choice chemistry nature of things objects that float or sink substances that sink or float 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?

  1. $0.01M$ $Ag{NO} _{3}$
  2. $0.04M$ ${K} _{2}{CO} _{3}$

  3. pure water
  4. in a buffer ($pH=4$)
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The solubility of Ag2CO3 is suppressed by the common ion effect. Ag+ from AgNO3 will significantly decrease the solubility of Ag2CO3 more than the other options provided.

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

$Ca _3(PO _4) _2$ 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?

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

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

  1. $6 \times 10^{-3}$M
  2. $10^{-6}$M
  3. $1.5 \times 10^{-4}$M
  4. $2.47 \times 10^{-9}$M
Reveal answer Fill a bubble to check yourself
D Correct answer
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}$

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

Which of the following is most soluble?

  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. ${Ag} _{2}S\left({K} _{sp} = 6\times {10}^{-51}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $

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

$PbCl _2$ 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 : 

  1. $4 \times 10^{-7}$M
  2. $4 \times 10^{-9}$M
  3. $2 \times 10^{-7}$M
  4. $2 \times 10^{-9}$M
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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

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