Tag: nature of things

Quizzes (10) Questions (75)

Questions Related to nature of things

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

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

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

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