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A 19.5-mL sample of Ba(OH)2 is titrated with HCl. If 28.4 mL of 0.394 M HCl is needed to reach the endpoint, what is the concentration (M) of the Ba(OH)2 solution

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The concentration of Ba(OH)₂ IS 0.287 M.

What is concentration?

This is the amount of solute in mol or in gram that will dissolve in 1 dm³ solution.

Equation of the reaction

  • Ba(OH)₂+2HCl→ BaCl₂+2H₂O

Formula:

  • CaVa/CbVb = Na/Nb................ Equation 1

Where:

  • Ca = Concentration of HCl
  • Va = Volume of HCl
  • Cb = Concentration of Ba(OH)₂
  • Vb = Volume of Ba(OH)₂
  • Na = number of moles of HCl
  • Nb = number of moles of Ba(OH)₂

Make Cb the subject of the equation

  • Cb = (CaVaNb)/(NaVb)................... Equation 2

From the equation and the question,

Given:

  • Ca = 0.394 M
  • Va = 28.4 mL
  • Vb = 19.5 mL
  • Na = 2
  • Nb = 1

Substitute these values into equation 2

  • Cb = (0.394×28.4×1)/(2×19.5)
  • Cb = 0.287 M.

Hence, The concentration of Ba(OH)₂ IS 0.287 M.

Learn more about concentration here: https://brainly.com/question/17206790


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