A saturated solution of M(OH) 2 is prepared and then filtered to remove any undi

Question

A saturated solution of M(OH)2 is prepared and then filtered to remove any undissolved solid M(OH)2. 15.0 mL of the filtered, saturated solution was titrated with a 0.060 M HCl solution. Examination of the pH titration curve indicated that it took 8.00 mL of the HCl solution to reach the equivalence point. Use this information to answer questions 12, 13 and 14.

12. What is the [OH-] in the saturated solution?

[OH-] = M

Tries 0/2

13. What is the [M2+] in the saturated solution?

[M2+] = M

Tries 0/2

14. What is the value of Ksp for M(OH)2?

Ksp =

Tries 0/2

Tries 0/2

Explanation / Answer

M(OH)2 (s)---> M2+ (aq) + 2OH- (aq)

This OH- reacts with HCl   to get neutralsied.

At equivalence point HCl moles = OH- moles = M x V = 0.06 x ( 8/1000) = 0.00048 = 4.8 x 10^ -4

Hence 15 ml of solution has 4.8 x 10^ -4 moles OH-

[OH-] = ( 4.8 x 10^ -4 ) / ( 0.015 = 0.032 M

13)   [Mn2+] = 1/2 ( [OH-]] = ( 1/2) 0.032 = 0.016

14) Ksp =[Mn2+][OH-]^2 = ( 0.016) ( 0.032)^2 = 1.6384 x 10^ -5

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