acid – base equilibrium problem #13. example: calculate the number of grams of nh 4 br that have...
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![Page 1: 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](https://reader037.vdocuments.mx/reader037/viewer/2022103123/56649d755503460f94a5687b/html5/thumbnails/1.jpg)
Acid – Base EquilibriumProblem #13
![Page 2: 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](https://reader037.vdocuments.mx/reader037/viewer/2022103123/56649d755503460f94a5687b/html5/thumbnails/2.jpg)
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.
![Page 3: 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](https://reader037.vdocuments.mx/reader037/viewer/2022103123/56649d755503460f94a5687b/html5/thumbnails/3.jpg)
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+,
![Page 4: 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](https://reader037.vdocuments.mx/reader037/viewer/2022103123/56649d755503460f94a5687b/html5/thumbnails/4.jpg)
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+,
![Page 5: 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](https://reader037.vdocuments.mx/reader037/viewer/2022103123/56649d755503460f94a5687b/html5/thumbnails/5.jpg)
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.
![Page 6: 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](https://reader037.vdocuments.mx/reader037/viewer/2022103123/56649d755503460f94a5687b/html5/thumbnails/6.jpg)
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.
![Page 7: 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](https://reader037.vdocuments.mx/reader037/viewer/2022103123/56649d755503460f94a5687b/html5/thumbnails/7.jpg)
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.
![Page 8: 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](https://reader037.vdocuments.mx/reader037/viewer/2022103123/56649d755503460f94a5687b/html5/thumbnails/8.jpg)
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.
![Page 9: 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](https://reader037.vdocuments.mx/reader037/viewer/2022103123/56649d755503460f94a5687b/html5/thumbnails/9.jpg)
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.