Question

If solubility product of base M(OH)3 is 2.7×10¹¹, the concentration of OH¹ will be

Answer 1

Answer: The answer is 3 × 10 ^-3

Explanation:

Answer 2

The concentration of hydroxide ion is 0.003 M

Explanation:

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The equation for the ionization of the metal (III) hydroxide is given as:

            $M(OH)_3\leftrightharpoons M^{3+}+3OH^-$

                                 s           3s

Expression for the solubility product of $M(OH)_3$ will be:

$K_{sp}=[M^{3+}][OH^-]^3\\\\K_{sp}=s\times (3s)^3=27s^4$

We are given:

$K_{sp}=2.7\times 10^{-11}$  

Putting values in above equation, we get:

$2.7\times 10^{11}=27s^4\\\\s=0.001M$

So, the concentration of $OH^-$ = 3s = (3 × 0.001) = 0.003 M

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