Question

oxalic acid (h₂c₂o₄) is a polyprotic acid. write balanced chemical equations for the sequence of reactions that oxalic acid can undergo when its dissolve water. h₂c₂o₄(aq) + h₂o(l) → h₃o⁺(aq) + hc₂o₄⁻(aq) hc₂o₄⁻(aq) + h₂o(l) → h₃o(aq) + c₂o₄²⁻(aq) h₂c₂o₄(aq) + naoh(aq) = nahc₂o(aq) nahc₂o(aq) + naoh(aq) = na₂c₂o₄(aq) be careful aleks does not understand your answer. • check your chemical symbols: did you capitalize by mistake or forget to capitalize? did you write 1 for lower l or upper case i, or a zero for capital o? • did you forget subscripts and/or superscripts? • check your parentheses. are they in matched pairs? if you are certain your answer contains no mistakes, please find an equivalent way of expressing it that aleks can understand. ok

Explanation:

Step1: First dissociation of oxalic acid

Oxalic acid (\(H_{2}C_{2}O_{4}\)) donates one proton (\(H^{+}\)) to water (\(H_{2}O\)) to form hydronium ion (\(H_{3}O^{+}\)) and hydrogen oxalate ion (\(HC_{2}O_{4}^{-}\)).

$$H_{2}C_{2}O_{4}(aq)+H_{2}O(l) ightleftharpoons H_{3}O^{+}(aq)+HC_{2}O_{4}^{-}(aq)$$

Step2: Second dissociation of oxalic acid

The hydrogen oxalate ion (\(HC_{2}O_{4}^{-}\)) donates another proton (\(H^{+}\)) to water (\(H_{2}O\)) to form hydronium ion (\(H_{3}O^{+}\)) and oxalate ion (\(C_{2}O_{4}^{2 -}\)).

$$HC_{2}O_{4}^{-}(aq)+H_{2}O(l) ightleftharpoons H_{3}O^{+}(aq)+C_{2}O_{4}^{2 -}(aq)$$

Answer:

$$H_{2}C_{2}O_{4}(aq)+H_{2}O(l) ightleftharpoons H_{3}O^{+}(aq)+HC_{2}O_{4}^{-}(aq)$$

$$HC_{2}O_{4}^{-}(aq)+H_{2}O(l) ightleftharpoons H_{3}O^{+}(aq)+C_{2}O_{4}^{2 -}(aq)$$