If pK b for CN- at 25° C is 4.7 , the pH of 0.5 M aqueous NaCN solution is:

Question

If pK b for CN- at 25° C is 4.7 , the pH of 0.5 M aqueous NaCN solution is: (A) 12 (B) 10 (C) 11.5 (D) 11

Tardigrade Admin · Accepted Answer

CN -+ H 2 O arrow HCN + OH - ; pK b =4.7 ∴ pK a =9.3 ( overset-O H )= ch = c √( K H / c )=√ K H ⋅ c =√( K w / K a ) ⋅ c =√(10-14 × 0.5/10-9.3) [ OH -]=√10-4.7 × 0.5 ∴ pOH =11.5

Q. If $p K_{b}$ for $C N^{-}$ at $2 5^{\circ} C$ is 4.7 , the $p H$ of $0.5 M$ aqueous $N a CN$ solution is:

12

10

11.5

11

$C N^{-} + H_{2} O \to H CN + O H^{-}; p K_{b} = 4.7$

$∴ p K_{a} = 9.3$

$(O - H) = c h = c \frac{K _{H}}{c} = K_{H} \cdot c = \frac{K _{w}}{K _{a}} \cdot c$

$= \frac{1 0 ^{- 14} \times 0.5}{1 0 ^{- 9.3}}$

$[O H^{-}] = 1 0^{- 4.7} \times 0.5$

$∴ pO H = 11.5$

Q. If pKb​ for CN− at 25∘C is 4.7 , the pH of 0.5M aqueous NaCN solution is:

12

10

11.5

11

CN−+H2​O→HCN+OH−;pKb​=4.7 ∴pKa​=9.3 (O−H)=ch=ccKH​​​=KH​⋅c​=Ka​Kw​​⋅c​ =10−9.310−14×0.5​​ [OH−]=10−4.7×0.5​ ∴pOH=11.5

Q. If $p K_{b}$ for $C N^{-}$ at $2 5^{\circ} C$ is 4.7 , the $p H$ of $0.5 M$ aqueous $N a CN$ solution is:

$C N^{-} + H_{2} O \to H CN + O H^{-}; p K_{b} = 4.7$

$∴ p K_{a} = 9.3$

$(O - H) = c h = c \frac{K _{H}}{c} = K_{H} \cdot c = \frac{K _{w}}{K _{a}} \cdot c$

$= \frac{1 0 ^{- 14} \times 0.5}{1 0 ^{- 9.3}}$

$[O H^{-}] = 1 0^{- 4.7} \times 0.5$

$∴ pO H = 11.5$