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Stroniyum fluoride (SrF(2.)) is a sparin...

Stroniyum fluoride `(SrF_(2.))` is a sparingly soluble slat. Let`s_(1)` be its solubility (in mol/lt.) in pure water at `25^@C`, assuming no hydrolysis of `F^(-)` ions. Also let `s_(2)` be its solubility (in hydrolysis of `F^(-)`ion and no complex formation. However,it is known that `s_(1):s_(2)=10^(6):256`.
The mass of NaF to be added to 100 ml solution of `0.0011MSr^(+2)` ions to reduce its concentration to `2xx10^(-4)`M is : [Assume no hydrolysis of `F^(-)` ions.

A

0.42 g

B

0.063g

C

0.021 g

D

0.084 g

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The correct Answer is:
c
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Stroniyum fluoride (SrF_(2.)) is a sparingly soluble slat. Let s_(1) be its solubility (in mol/lt.) in pure water at 25^@C , assuming no hydrolysis of F^(-) ions. Also let s_(2) be its solubility (in hydrolysis of F^(-) ion and no complex formation. However,it is known that s_(1):s_(2)=10^(6):256 . The K_(sp) value of SrF_(2) at 25^(@) C is :

Stroniyum fluoride (SrF_(2.)) is a sparingly soluble slat. Let s_(1) be its solubility (in mol/lt.) in pure water at 25^@C , assuming no hydrolysis of F^(-) ions. Also let s_(2) be its solubility (in hydrolysis of F^(-) ion and no complex formation. However,it is known that s_(1):s_(2)=10^(6):256 . The solubility of SrF_(2) (in mol/L) in a buffer solution of pH=5 at 25^(@) C is :

K_(sp) of SrF_(2) = 2.8 xx 10^(-9) at 25^(@)C . How much NaF should be added to 100mL of solution having 0.016M in Sr^(2+) ions to reduce its concentration to 2.5 xx 10^(-3)M ?

How much NaF should be added to 100mL of solution having 0.016M in Sr^(2+) ions to reduces its concentration to 2.5 xx 10^(-3)M? K_(sp) SrF_(2) = 2.8 xx 10^(-9)at 298K .

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