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The pH in aqueous solution of hcl at 25c is 4 hcl0.003 moles was added to 1l of this solution calculate the new pH in the solution assume complete dissociation of hcl?
Wiki User
∙ 15y ago
if its complete dissociation, then the products would be a salt and water, which means the pH is 7 or neutral. OMG, if the pH is currently 4 then [H+] = 1.0 e-4 M (pH = -log[H+]) if you add 0.003 moles then 1.0e-4 M +.003 M = .0031 M (Since the strong acid HCL completely dissociates in aq solution) pH = -log [.0031M] = 2.51
Wiki User
∙ 15y ago
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