0
You visited us 0 times! Enjoying our articles? Unlock Full Access!
Question

The $$K_{sp}$$ of $$Mg(OH)_{2}$$ is $$1\times 10^{-12}, 0.01\ M\ Mg(OH)_{2}$$ will precipitate at the limiting $$pH$$

A
$$3$$
B
$$9$$
C
$$8$$
D
$$5$$
Solution
Verified by Toppr

Correct option is B. $$9$$
$$Mg(OH)_{2}\rightleftharpoons Mg^{2+}+2OH^{-}$$'

$$K_{sp}=[Mg^{2+}][OH^{-}]^{2}$$

$$1\times 10^{-12}=0.01\times [OH^{-}]^{2}$$

$$[OH^{-}]^{2}=1\times 10^{-10}\Rightarrow [OH^{-}]=10^{-5}$$

$$[H^{+}]=\dfrac{10^{-14}}{10^{-5}}=10^{9}$$ $$\because k_w=[H^+][OH^-]$$

$$pH=-\log [H^{+}]=-\log [10^{9}]=9$$

Was this answer helpful?
0
Similar Questions
Q1
The $$K_{sp}$$ of $$Mg(OH)_{2}$$ is $$1\times 10^{-12}, 0.01\ M\ Mg(OH)_{2}$$ will precipitate at the limiting $$pH$$
View Solution
Q2
The Ksp of Mg(OH)2 is 1×1012. 0.01 m Mg(OH)2 will precipitate at the limiting pH equal to:
View Solution
Q3
KspofMg(OH)2 is 1×1012,0.01MMgCl2 will be precipitating at the limiting pH :
View Solution
Q4
Calculate the pH at which $$Mg(OH)_2$$ begins to precipitate from a solution containing $$0.1$$ M $$Mg^{2+}$$ ions, $$K_{sp}$$ for $$Mg(OH)_2=1\times 10^{-11}$$.
View Solution
Q5
What is the maximum pH of a solution of $$0.1$$M in $$Mg^{2+}$$ from which $$Mg(OH)_2$$ will not precipitate $$K_{sp}[Mg(OH)_2]=1.2\times 10^{-11}$$.
View Solution