If 1.2 mL of 0.01 M HCl is added to a 350 mL solution containing 0.500 M succinic acid and 0.200 M potassium succinate, what is the resulting pH of the solution? (Identify the acid and the conjugate base, write out all equations). Ka succinic acid= 6.17 x 10^-5

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Question:

If 1.2 mL of 0.01 M HCl is added to a 350 mL solution containing 0.500 M succinic acid and 0.200 M potassium succinate, what is the resulting pH of the solution? (Identify the acid and the conjugate base, write out all equations). Ka succinic acid= 6.17 x 10^-5

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Answer:pH is 3,81Explanation:The equilibrium: Succinic acid ⇄ Succinate + H⁺; ka = 6,17x10⁻⁵The initial moles of each compound are:Succinic acid: 0,350L × 0,500M = 0,175 molesSuccinate: 0,350L × 0,200M = 0,07 molesThe addition of HCl will consume succinate and produce succinic acid, thus:Moles of HCl:0,0012L × 0,01M = 1,2x10⁻⁵ molesSuccinic acid: 0,175 moles +  1,2x10⁻⁵ moles = 0,175012 molesSuccinate: 0,07 moles -  1,2x10⁻⁵ moles = 0,069988 molesThe equilibrium equation is:ka = [tex]\frac{[succinate][H^{+}]}{[Succinic Acid]}[/tex]Replacing:6,17x10⁻⁵ = [tex]\frac{[0,069988][H^{+}]}{[0,175012]}[/tex]Solving:[H⁺] = 1,543x10⁻⁴As pH = -log [H⁺]pH = 3,81I hope it helps!

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