What will happen if 0.001 M Mg(no3)2 solution is adusted to ph=9 ksp of mg(no3)2 is 8.9*10^-12
Answers
Explanation:
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Answer:
No ppt.
Explanation:
We get from the question that the solution's pH 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.