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0.125 mole of CaA2 are present in 0.5litres solution if Ka of HA is 8.0x10^-4 and salt CaA2 is completely ionized thr ph of solutions is?
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0.125 mole of CaA2 are present in 0.5litres solution if Ka of HA is 8....
Calculation of pH of Solution Containing 0.125 mole of CaA2

Given:

  • Amount of CaA2 = 0.125 mole

  • Volume of solution = 0.5 litres

  • Ka of HA = 8.0 x 10^-4



Step 1: Calculate the concentration of CaA2 in the solution.

  • Moles of CaA2 = 0.125 mole

  • Volume of solution = 0.5 litres

  • Concentration of CaA2 = Moles of CaA2 / Volume of solution

  • Concentration of CaA2 = 0.125 mole / 0.5 litres

  • Concentration of CaA2 = 0.25 M



Step 2: Calculate the concentration of H+ ions in the solution using the dissociation constant of HA.

  • Ka of HA = 8.0 x 10^-4

  • HA + H2O ⇌ H3O+ + A-

  • Let x be the concentration of H3O+ ions produced when the HA dissociates.

  • Ka = [H3O+][A-] / [HA]

  • 8.0 x 10^-4 = x^2 / (0.25 - x)

  • Assuming x < 0.25,="" simplify="" the="" equation="" />

  • 8.0 x 10^-4 = x^2 / 0.25

  • x = 0.02 M

  • Therefore, the concentration of H+ ions in the solution is 0.02 M



Step 3: Calculate the pH of the solution.

  • pH = -log[H+]

  • pH = -log(0.02)

  • pH = 1.7



Therefore, the pH of the solution containing 0.125 mole of CaA2 is 1.7.
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0.125 mole of CaA2 are present in 0.5litres solution if Ka of HA is 8.0x10^-4 and salt CaA2 is completely ionized thr ph of solutions is?
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