What is the concentration (in M) of the hydroxide ion when 18.6 mL of a 0.538 M solution of sodium hydroxide i

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We must assume that both are strong electrolytes. That means that the salts are 100% dissociated in aqueous solution.

First calculate the moles of OH^- in each solution. Then add the moles of each part to find the total moles of OH^-. Finally, calculate the molarity of the OH^- ion.

In NaOH solution: NaOH(s) in water --> Na^+(aq) + OH^-(aq)

moles NaOH = (Vol in L)(molarity) = (0.0186 L)(0.538 mol/L) = 0.0100068 moles NaOH

Since 1 mole of NaOH gives 1 mole of OH^- in solution then you have 0.0100068 moles OH^- from NaOH. Moles OH^- = 0.0100068 moles

In Ba(OH)2 solution: Ba(OH)2(s) in water --> Ba^2+(aq) + 2 OH^-(aq)

moles Ba(OH)2 = (0.0168 L)(0.509 mol/L) = 0.0085512 moles Ba(OH)2

Since 1 moles of Ba(OH)2 yields 2 moles of OH^- then you have 2(0.0085512 moles OH^-) = 0.0171024 moles OH^-

Total number of moles of OH^- in solution = 0.0100068 moles + 0.0171024 moles = 0.0271092 moles

molarity = moles of ions/Liters of solution = (0.0271092 mol)/(0.0186 L + 0.0168 L) = 0.766 M OH^-

Hope this is helpful to you. JIL HIR

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