Ka for HF is 3-5- 10-4- Calculate Kb for the fluoride ion-

The correct option is C 2.9× 10^ 11 K_a = 3.5 × 10^ 4 K_b #160; for fluoride ion wills be : K_b= K_wK_a=1.0 × 10^ 143.5 × 10^ 4=2.9 × 10^ ...

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Standard XII

Chemistry

Auto Protolysis of Water

Ka for HF i...

Question

$K_{a}$ for $H F$ is $3.5 \times 10^{- 4}$ . Calculate $K_{b}$ for the fluoride ion.

3.5 \times 10^{- 4}

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1.0 \times 10^{- 7}

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2.9 \times 10^{- 11}

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1.0 \times 10^{- 14}

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Solution

The correct option is C $2.9 \times 10^{- 11}$
$K_{a} = 3.5 \times 10^{- 4}$

$K_{b}$ for fluoride ion wills be :

$K_{b} = \frac{K_{w}}{K_{a}} = \frac{1.0 \times 10^{- 14}}{3.5 \times 10^{- 4}} = 2.9 \times 10^{- 11}$

$∴$ correct option is $(C)$

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Similar questions

Q. What is the

H^{+}

ion concentration in ammonium acetate solution?

(K_{a} = 1.8 \times 10^{- 5}, K_{b} = 1.8 \times 10^{- 5}, K_{w} = 1.0 \times 10^{- 14})

A certain weak acid has $K a = 1.0 \times 10^{- 4}$ . Calculate the equilibrium constant for its reaction with a strong Base.

Substance	$k_{a}$
Hydrogen carbonate ion (bicarbonate ion) $H C O_{3}^{-}$	$4.8 \times 10^{- 11}$
Hydrogen phosphate ion $H P O_{4}^{- 3}$	$3.6 \times 10^{- 13}$
Water $H_{2} O$	$1.0 \times 10^{- 14}$
Hydrogen sulfide ion $H S^{-}$	$1.0 \times 10^{- 19}$

Using acid dissociation constants above, which substance is the strongest base?

Q. Calculate the pH of

1.0 \times 10^{- 3}

M sodium phenolate,

N a O C_{6} H_{5} \cdot K_{a}

for

H O C_{6} H_{5}

0.6 \times 10^{- 10}

Calculate the emf of the cell

Pt, $H_{2} (1.0 a t m)$ | $C H_{3} C O O H$ (0.1M) || $N H_{3}$ (aq, 0.001M) | $H_{2}$ (1.0 atm), Pt Ka( $C H_{3} C O O H$ ) = $10^{- 5}$ , Kb(NH3) = $10^{- 5}$

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Ka for HF is 3.5×10−4. Calculate Kb for the fluoride ion.

The correct option is C 2.9×10−11Ka=3.5×10−4Kb for fluoride ion wills be :Kb=KwKa=1.0×10−143.5×10−4=2.9×10−11∴ correct option is (C)

$K_{a}$ for $H F$ is $3.5 \times 10^{- 4}$ . Calculate $K_{b}$ for the fluoride ion.

The correct option is C $2.9 \times 10^{- 11}$
$K_{a} = 3.5 \times 10^{- 4}$

$K_{b}$ for fluoride ion wills be :

$K_{b} = \frac{K_{w}}{K_{a}} = \frac{1.0 \times 10^{- 14}}{3.5 \times 10^{- 4}} = 2.9 \times 10^{- 11}$

$∴$ correct option is $(C)$