Question

At 25 oC the solubility of iron(II) hydroxide is 1.59 x 10-5 mol/L. Calculate the value of Ksp at this temperature. Give your answer in scientific notation to 2 SIGNIFICANT FIGURES (even though this is strictly incorrect). [a]

Answer 1

Answer:

$Ksp=1.4\times 10^{-14}$

Explanation:

 Balance equaion of the given reacion is:

$Fe(OH)_2$ ⇄ $Fe^{2+} +2OH^-$

lets solubility is S

given:

$S=1.59\times10^{-5} mol/L$

Solubility of each ion;

$[Fe^{2+}]= S;$

$[OH^-]=2S;$

$Ksp=S\times(2S)^2;$

$Ksp=4S^3$

after putting value of S the value of Ksp is:

$Ksp=4\times(1.5\times10^{-5} )^{3}$

$Ksp=1.35\times 10^{-14}$

Upto 2 significant figure:

$Ksp=1.4\times 10^{-14}$

Answer 2

The Ksp of the solution is $1.607*10^-^1^4$

Data;

• Temperature = 25°
• solubility = 1.59*10^-5 mol/L
• Ksp = ?

Solubility Constant

This is the point or temperature in which a solute is completely soluble in a solvent.

The solubility of iron(ii) hydroxide can be calculated by using the equation of reaction.

$Fe(OH)_2 \to Fe^2^+ + 2OH^-$

But the solubility of the solution is given as 1.59*10^-5 mol/L

$[Fe^2^+] = s$

$[OH^-] = 2s\\K_s_p = s * (2s)^2 = 4s^3$

Let's substitute the values and solve

$K_s_p = 4 *( 1.59*10^-^5)^3 = 1.607*10^-^1^4$

The Ksp of the solution is $1.607*10^-^1^4$

Learn more on Ksp of a solution here;

https://brainly.com/question/10710012