The ionization constant of HF is 7.0 ×10-4M. The pH of 0.l M NaF solution will be

F−+H2O⇌HF+OH− c - x x xKb=[HF]OH−F−=x2c−x≃x2c; ( assuming x « c) orOH−=x=Kbc=Kw/Kac=1.0×10−14M2/7.0×10−4M(0.1M)1/2=1.2×10−6M pOH =−log⁡1.2×10−6=5.92; pH =14−pOH=8.08

The ionization constant of HF is 7.0 ×10-4M. The pH of 0.l M NaF solution will be

F−+H2O⇌HF+OH− c - x x xKb=[HF]OH−F−=x2c−x≃x2c; ( assuming x « c) orOH−=x=Kbc=Kw/Kac=1.0×10−14M2/7.0×10−4M(0.1M)1/2=1.2×10−6M pOH =−log⁡1.2×10−6=5.92; pH =14−pOH=8.08

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The ionization constant of HF is $7.0 \times 10^{- 4} M$ . The pH of 0.l M NaF solution will be

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Detailed Solution

$F^{-} + H_{2} O ⇌ HF + {OH}^{-} c - x x x$

$K_{b} = \frac{[HF] [{OH}^{-}]}{[F^{-}]} = \frac{x^{2}}{c - x} ≃ \frac{x^{2}}{c};$ ( assuming x « c) or $[{OH}^{-}] = x = \sqrt{K_{b} c} = \sqrt{(K_{w} / K_{a}) c} = {[(1.0 \times 10^{- 14} M^{2} / 7.0 \times 10^{- 4} M) (0.1 M)]}^{1 / 2} = 1.2 \times 10^{- 6} M$ $pOH = - \log (1.2 \times 10^{- 6}) = 5.92; pH = 14 - pOH = 8.08$