Given NO−2+H2O⇋HNO2+OH−, Ka(HNO2)=7.2×10−4 Thus, the degree – InterviewSolution

1.	Given NO−2+H2O⇋HNO2+OH−, Ka(HNO2)=7.2×10−4 Thus, the degree of hydrolysis of 0.05 M NaNO2 solution is:
Answer» Given $N O_{2}^{-} + H_{2} O ⇋ H N O_{2} + O H^{-}$ , $K_{a} (H N O_{2}) = 7.2 \times 10^{- 4}$ Thus, the degree of hydrolysis of 0.05 $M$ $N a N O_{2}$ solution is:

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