Question

Find the literature ksp value for Mn(OH )2 . ph of saturated solution mn(Oh )2 is 9.90 calculate oh ,mn, mnoh2 concentration and , ksp .compare calculated ksp value from the literature ksp value . are the two values close or different .

Answer 1

Answer:-

1. $Literature K_{sp} \approx 1.9 \times 10^{-13}$ and  $K_{sp} = 2.5 \times 10^{-13}$

both K_sp values are close to each other

2. $[Mn^{+}] = 3.97 \times 10^{-5} M$

3. $[OH^{-}] = 7.94 \times 10^{-5} M$

4. $[Mn(OH)_{2}] = 3.97 \times 10^{-5} M$

Exlanation :-

The solubility equilibrium of Mn(OH)₂ is

$Mn(OH)_{2}_{(s)} \rightleftharpoons Mn^{2+}_{(aq)} + 2OH^{-}_{(aq)}$ ...(1)

The K_sp of Mn(OH)₂ is given as

$K_{sp} = [Mn^{2+}][OH^{-}]^{2}$ ....(2)

Let

the molar solubility of Mn(OH)₂ in the solution be S (mol/L).

therefore,

$[Mn^{2+}] = S (mol/L)$ and  $[OH^{-}] = 2S (mol/L)$

therefore, $K_{sp} = S \times (2S)^{2}$ ...(3)

We are given p^H = 9.9

therefore,

p^OH = 14 - p^H = (14 - 9.9) = 4.1

Now,

$p^{OH} = - log[OH^{-}]$

ie. $- log[OH^{-}] = 4.1$

i.e $log[OH^{-}] = - 4.1$

ie. $[OH^{-}] = antilog (- 4.1)$

i.e $[OH^{-}] = 7.94 \times 10^{-5} M$

ie. $[OH^{-}] = 2S = 7.94 \times 10^{-5} M$

from reaction (1) we have,

$[Mn^{+}] = S = \frac{[OH^{-}]}{2} = \frac{7.94 \times 10^{-5} M}{2} = 3.97 \times 10^{-5} M$

Therefore,

from eq(3) we have,

$K_{sp} = 3.97 \times 10^{-5} \times (7.94 \times 10^{-5})^{2}$

$K_{sp} = 3.97 \times 10^{-5} \times 63.0436 \times 10^{-10}$

$K_{sp} = 2.50 \times 10^{-13}$

$Literature K_{sp} \approx 1.9 \times 10^{-13}$

Now,

from, eq(3)

$K_{sp} = S \times (2S)^{2}$  

$K_{sp} = 4(S)^{3}$

$(S)^{3} = \frac{ K_{sp}}{4} = \frac{2.50 \times 10^{-13}}{4}$

$(S)^{3} = 6.25 \times 10^{-14}$

taking cube root of both side we get,

$S = 3.97 \times 10^{-5}$

i.e $[Mn(OH)_{2}] = 3.97 \times 10^{-5}$