A saturated solution of manganese(II) hydroxide was prepared, and an acid–base titration was performed to determine its KspKsp at 25 °C. The endpoint was reached when 70.00 mLmL of the manganese(II) hydroxide solution was titrated with 4.86 mLmL of 0.0020 MM HClHCl solution. What is the KspKsp of manganese(II) hydroxide?

Respuesta :

Answer:

10.945 x 10^-4

Explanation:

Balanced equation:

Mn(OH)2 +  2 HCl  -->  MnCl2  +  H2O

it takes 2 moles HCL for each mole Mn(OH)2

Next find the molarity of the Mn(OH)2 solution

= (1 mole Mn(OH)2 / 2 mole HCl)  X (0.0020 mole HCl / 1000ml) X (4.86 ml)    

= 4.86 x 10^-3 mole  

this is now dissolved in (70 + 4.86)  =  74.86 ml or 0.07486 L

thus [Mn(OH)2]  =  4.86 x 10^-3 mole / 0.07486 L  =  0.064921 M

Ksp =  [Mn2+][OH-]^2  =  4x^3  =  4(0.064921)^3  = 10.945 x 10^-4