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11 Nov 2019
Help would be great Hydrosulfuric acid (H_2S) is considered a 'weak acid." It dissociates in water according to this reaction: H_2 S(aq) + H_2O(t) HS^-(aq) + H_3O^+(aq) k_a = 9 times 10^-8 Hydroiodic acid (HI) is considered a "strong acid". It dissociates in water according to this reaction: HI(aq) + H_2O(l) rightarrow I^-(aq) + H_3O^+(aq) K_a = 3 times 10^9 Suppose you have two beakers that each contain 750 mL of water. In one, you dissolve 0.30 moles H_2S. In the other, you dissolve 0.30 moles HI. Without doing any calculations, draw graphs that qualitatively represent the initial and equilibrium concentrations of reactants and products in each beaker. (Either type of graph you drew in Question 5 from Recitation 1 would be appropriate.) Explain why you drew your graphs in this way. Calculate the equilibrium concentration of hydronium ion (H_3O^+) in the beaker containing the weak acid, H_2S.
Help would be great
Hydrosulfuric acid (H_2S) is considered a 'weak acid." It dissociates in water according to this reaction: H_2 S(aq) + H_2O(t) HS^-(aq) + H_3O^+(aq) k_a = 9 times 10^-8 Hydroiodic acid (HI) is considered a "strong acid". It dissociates in water according to this reaction: HI(aq) + H_2O(l) rightarrow I^-(aq) + H_3O^+(aq) K_a = 3 times 10^9 Suppose you have two beakers that each contain 750 mL of water. In one, you dissolve 0.30 moles H_2S. In the other, you dissolve 0.30 moles HI. Without doing any calculations, draw graphs that qualitatively represent the initial and equilibrium concentrations of reactants and products in each beaker. (Either type of graph you drew in Question 5 from Recitation 1 would be appropriate.) Explain why you drew your graphs in this way. Calculate the equilibrium concentration of hydronium ion (H_3O^+) in the beaker containing the weak acid, H_2S.
Jarrod RobelLv2
22 Jun 2019