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A student is asked to standardize a solution of barium hydroxide. He weighs out 0.978 g potassium hydrogen phthalate (KHC8H4O4, treat this as a monoprotic acid). It requires 35.8 mL of barium hydroxide to reach the endpoint. A. What is the molarity of the barium hydroxide solution? M This barium hydroxide solution is then used to titrate an unknown solution of hydrochloric acid. B. If 17.1 mL of the barium hydroxide solution is required to neutralize 18.6 mL of hydrochloric acid, what is the molarity of the hydrochloric acid solution? M

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Answer:

(A) 0.129 M

(B) 0.237 M

Explanation:

(A) The reaction between potassium hydrogen phthalate and barium hydroxide is:

  • 2HA + Ba(OH)₂ → BaA₂ + 2H₂O

Where A⁻ is the respective anion of the monoprotic acid (KC₈H₄O₄⁻).

We convert mass of phthalate to moles, using its molar mass:

  • 0.978 g ÷ 156 g/mol = 9.27x10⁻³ mol = 9.27 mmol

Now we convert mmol of HA to mmol of Ba(OH)₂:

  • 9.27 mmol HA * [tex]\frac{1mmolBa(OH)_{2}}{2mmolHA}[/tex] = 6.64 mmol Ba(OH)₂

Finally we calculate the molarity of the Ba(OH)₂ solution:

  • 6.64 mmol / 35.8 mL = 0.129 M

(B) The reaction between Ba(OH)₂ and HCl is:

  • 2HCl + Ba(OH)₂ → BaCl₂ + 2H₂O

So the moles of HCl that reacted are:

  • 17.1 mL * 0.129 M * [tex]\frac{2mmolHCl}{1mmolBa(OH)_2}[/tex] = 4.41 mmol HCl

And the molarity of the HCl solution is:

  • 4.41 mmol / 18.6 mL = 0.237 M

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