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Acid – Base Equilibrium Problem #13

Acid – Base Equilibrium Problem #13. Example: Calculate the number of grams of NH 4 Br that have to be dissolved in 1.00L of water at 25 o C to have a solution with a pH of 5.1.

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Acid – Base Equilibrium Problem #13

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  1. Acid – Base EquilibriumProblem #13

  2. Example: Calculate the number of grams of NH4Br that have to be dissolved in 1.00L of water at 25oC to have a solution with a pH of 5.1.

  3. Example: Calculate the number of grams of NH4Br that have to be dissolved in 1.00L of water at 25oC to have a solution with a pH of 5.1. The only dissociation that we need to consider is: NH4+ NH3 + H+,

  4. Example: Calculate the number of grams of NH4Br that have to be dissolved in 1.00L of water at 25oC to have a solution with a pH of 5.1. The only dissociation that we need to consider is: NH4+ NH3 + H+,

  5. Example: Calculate the number of grams of NH4Br that have to be dissolved in 1.00L of water at 25oC to have a solution with a pH of 5.1.

  6. Example: Calculate the number of grams of NH4Br that have to be dissolved in 1.00L of water at 25oC to have a solution with a pH of 5.1.

  7. Example: Calculate the number of grams of NH4Br that have to be dissolved in 1.00L of water at 25oC to have a solution with a pH of 5.1.

  8. Example: Calculate the number of grams of NH4Br that have to be dissolved in 1.00L of water at 25oC to have a solution with a pH of 5.1.

  9. Example: Calculate the number of grams of NH4Br that have to be dissolved in 1.00L of water at 25oC to have a solution with a pH of 5.1. The only dissociation that we need to consider is: NH4+ NH3 + H+, if pH = 5.16, then [H+] = 10–pH = 10–5.16 = 6.9  10–6 M This is also the required value for [NH3], since they are formed in a one-to-one mole ratio. Ka= [H+][NH3]/[NH4+] = 5.6  10–10 and we arrive at: 5.6  10–10 = (6.9  10–6)(6.9  10–6)/[NH4+] [NH4+] = 8.5  10–2 M 8.5  10–2mol/L  97.9 g/mol = 8.3 g NH4Br are needed per liter.

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