What is the molar solubility of MnOH2 ksp - 4-5 - 10-14 in a buffer solution containing equal amounts of NH4- and NH3 kb - 1-8 - 10-5-

The correct option is B 1.38 × 10^ 4 For basic buffer solution #160; pOH=pK_a+logNH^+_4NH_3=pK_b ∴ [OH^ ]=1.8 × 10^ 5 K_sp=[MN^2+][OH^ ]^2 ...

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Byju's Answer

Standard XII

Chemistry

Basic Buffer Action

What is the m...

Question

What is the molar solubility of $M n (O H)_{2}$ ( $k_{s p}$ = 4.5 $\times 10^{- 14}$ ) in a buffer solution containing equal amounts of ${N H}_{4}^{+}$ and $N H_{3}$ ( $k_{b} = 1.8 \times 10^{- 5}$ )?

3.0

\times 10^{- 4}

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1.38

\times 10^{- 4}

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1.38

\times 10^{- 3}

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1.3

\times 10^{- 4}

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Solution

The correct option is B 1.38 $\times 10^{- 4}$
For basic buffer solution
$p O H = p K_{a} + l o g \frac{N H_{4}^{+}}{N H_{3}} = p K_{b}$
$∴ [O H^{-}] = 1.8 \times 10^{- 5}$
$K_{s p} = [M N^{2 +}] [O H^{-}]^{2}$
$[M n^{2 +}] = 1.38 \times 10^{- 4} M$

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Similar questions

The solubility of

M g {(O H)}_{2}

is increased by the addition of

{N H}_{4}^{+}

ion.

Given:

K_{s p}

M g (O H)_{2} = 6 \times 10^{- 12}

K_{b}

N H_{3} = 1.8 \times 10^{- 5}

Solubility of

M g {(O H)}_{2}

in a solution containing

0.5 M

{N H}_{4} C l

before addition of

M g {(O H)}_{2}

is:

Q. What is

[{N H}_{4}^{+}]

in a solution containing 0.02M

{N H}_{3}

(

K_{b} = {1.8 \times 10}^{- 5}

) and 0.01M KOH?

Q. The solubility of

C a F_{2}

in a solution buffered at

p H = 3.0

is........

K_{a}

for

H F = 6.3 \times 10^{- 4}

and

K_{s p}

C a F_{2} = 3.45 \times 10^{- 11}

Q. What is the solubility of

A l (O H)_{3}, (K_{s p} = 1 \times 10^{- 33})

in a buffer solution of

p H = 4

Q. The solubility of

C a F_{2}

in a solution buffered at

p H = 3.0

K_{a}

for

H F

6.3 \times 10^{- 4}

and

K_{s p}

C a F_{2} = 3.45 \times 10^{- 11} M

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What is the molar solubility of Mn(OH)2 (ksp = 4.5 ×10−14) in a buffer solution containing equal amounts of NH+4 and NH3 (kb=1.8×10−5)?

The correct option is B 1.38 ×10−4For basic buffer solution pOH=pKa+logNH+4NH3=pKb∴[OH−]=1.8×10−5Ksp=[MN2+][OH−]2[Mn2+]=1.38×10−4M

What is the molar solubility of $M n (O H)_{2}$ ( $k_{s p}$ = 4.5 $\times 10^{- 14}$ ) in a buffer solution containing equal amounts of ${N H}_{4}^{+}$ and $N H_{3}$ ( $k_{b} = 1.8 \times 10^{- 5}$ )?

The correct option is B 1.38 $\times 10^{- 4}$
For basic buffer solution
$p O H = p K_{a} + l o g \frac{N H_{4}^{+}}{N H_{3}} = p K_{b}$
$∴ [O H^{-}] = 1.8 \times 10^{- 5}$
$K_{s p} = [M N^{2 +}] [O H^{-}]^{2}$
$[M n^{2 +}] = 1.38 \times 10^{- 4} M$