acid – base equilibrium problem #13. example: calculate the number of grams of nh 4 br that have...
TRANSCRIPT
Acid – Base EquilibriumProblem #13
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.
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+,
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+,
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.
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.
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.
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.
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 MThis 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 M8.5 10–2 mol/L 97.9 g/mol = 8.3 g NH4Br are needed per liter.