Question

If ${pK}_b$ for fluoride ion at 25 is 10.83, the ionisation constant of hydrofluoric acid in water at this temperature is :

• A. $1.74\times {10}^{-5}$
• B. $3.52\times {10}^{-3}$
• C. $6.75\times {10}^{-4}$
• D. $5.38\times {10}^{-2}$

Answer 1

Answer: (C)

$6.75\times {10}^{-4}$

${pK}_b$ of fluoride ion is $10.83$

${pK}_b=10.83$

$K_b={10}^{-10.83}=1.48\times {10}^{-11}$

$HF+H_{2}O\rightleftharpoons F^{-}+H_3{O}^{+}$

$\left(K_a\right)\left(K_b\right)=K_w$

$K_a=\frac{K_w}{K_b}=\frac{{10}^{-14}}{1.48\times {10}^{-11}}=6.75\times {10}^{-4}$

$\Rightarrow$ ionisation constant of $HF$ is $6.75\times {10}^{-4}$.