10mL of 10-4M CaCl2 are mixed with 10mL of 10-4M NaF-Given K sp Ca F 2 -1-6-10-10-What is the product of ionic conc- for CaF2-

The correct option is A 1.25×10^ 13 [Ca^+2] = [F^ ] =10^ 4/2 = 5 × 10^ 5 The ionic product is [Ca^2+][F^ ]^2 = (5 × 10^ 5)(5 × 10^ 5)^2 =1.255 × ...

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Standard XII

Chemistry

Solubility Product

10mL of 10-...

Question

$10 m L$ of $10^{- 4} M$ $C a {C l}_{2}$ are mixed with $10 m L$ of $10^{- 4} M$ $N a F$ .
Given $K_{{s p}_{C a F_{2}}} = 1.6 \times 10^{- 10}$ .

What is the product of ionic conc. for $C a F_{2}$ ?

1.05 \times 10^{- 13}

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1.25 \times 10^{- 13}

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1.55 \times 10^{- 13}

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None of these

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Solution

The correct option is A $1.25 \times 10^{- 13}$
$[C a^{+ 2}] = [F^{-}] = \frac{10^{- 4}}{2} = 5 \times 10^{- 5}$
The ionic product is $[C a^{2 +}] [F^{-}]^{2} = (5 \times 10^{- 5}) (5 \times 10^{- 5})^{2} = 1.255 \times 10^{- 13}$ .

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10mL of 10−4M CaCl2 are mixed with 10mL of 10−4M NaF.Given KspCaF2=1.6×10−10.What is the product of ionic conc. for CaF2?

The correct option is A 1.25×10−13[Ca+2]=[F−]=10−42=5×10−5The ionic product is [Ca2+][F−]2=(5×10−5)(5×10−5)2=1.255×10−13.

$10 m L$ of $10^{- 4} M$ $C a {C l}_{2}$ are mixed with $10 m L$ of $10^{- 4} M$ $N a F$ .
Given $K_{{s p}_{C a F_{2}}} = 1.6 \times 10^{- 10}$ .

What is the product of ionic conc. for $C a F_{2}$ ?

The correct option is A $1.25 \times 10^{- 13}$
$[C a^{+ 2}] = [F^{-}] = \frac{10^{- 4}}{2} = 5 \times 10^{- 5}$
The ionic product is $[C a^{2 +}] [F^{-}]^{2} = (5 \times 10^{- 5}) (5 \times 10^{- 5})^{2} = 1.255 \times 10^{- 13}$ .