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Question: If Kb for fluoride ion at 25oC 1.48 x 10-11 the ionization constant of HF in water at this temperatu...

If Kb for fluoride ion at 25oC 1.48 x 10-11 the ionization constant of HF in water at this temperature is

A

1.7×10−51.7 \times 10^{- 5}

B

3.52×10−33.52 \times 10^{- 3}

C

6.75×10−46.75 \times 10^{- 4}

D

5.38×10−25.38 \times 10^{- 2}

Answer

6.75×10−46.75 \times 10^{- 4}

Explanation

Solution

Ka×kb=kω⇒Ka=KωKb=10−14148×10−11=6.75×10−4K_{a} \times k_{b} = k_{\omega} \Rightarrow K_{a} = \frac{K_{\omega}}{K_{b}} = \frac{10^{- 14}}{148 \times 10^{- 11}} = 6.75 \times 10^{- 4}