Question

What will happen if the ph of the solution of 0.001M Mg(NO3)2 solution is adjusted to ph=9?(Ksp Mg(OH) 2 = 8.9×10^-12)

Answer 1

Answer:

No ppt.

Explanation:

Since, from the question we get that the pH of the solution is adjusted to 9. Hence, the pOH will be 14-9 which will be 5. So, the concentration of the [OH]- ion will be 10^-5.

Since, the concentration of the Mg(NO3)2 is given in the question as 0.001 M. From this we get that the ionic concentration will be [Mg2+] =0.001M. Hence, the ionic product which is given as Mg(OH)2 will be 0.001*(10^-5)^2 which on solving we will get that 10^-13. Since, this result is smaller than the given ksp value of Mg(OH)2 which was 8.9×10^-12. So, there will not be any precipitate at this pH.  

Answer 2

Answer: no ppt

Explanation: