calculate the solubility of potassium bromide at 23 degrees C. HInt: Assume that the solubility increases by an equal amount for each degree between 20 degrees C and 30 degrees C.
The correct answer and explanation is :
To calculate the solubility of potassium bromide (KBr) at 23°C, we need reference values for the solubility at 20°C and 30°C. These values are commonly found in solubility charts.
From a standard solubility chart:
- At 20°C, the solubility of KBr is approximately 65 g/100 g of water.
- At 30°C, the solubility of KBr is approximately 74 g/100 g of water.
Step 1: Determine the total increase in solubility from 20°C to 30°C
$$
74\,g – 65\,g = 9\,g
$$
So, over a 10°C interval (from 20°C to 30°C), the solubility increases by 9 g.
Step 2: Determine the increase in solubility per 1°C
$$
\frac{9\,g}{10^\circ C} = 0.9\,g/^\circ C
$$
Step 3: Determine the increase from 20°C to 23°C (which is 3°C)
$$
0.9\,g/^\circ C \times 3^\circ C = 2.7\,g
$$
Step 4: Add the increase to the solubility at 20°C
$$
65\,g + 2.7\,g = 67.7\,g/100\,g\,\text{of water}
$$
✅ Final Answer: 67.7 g of KBr per 100 g of water at 23°C
✏️ Detailed Explanation (300+ words):
Solubility refers to the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature to form a saturated solution. In this case, we’re calculating how much potassium bromide (KBr), an ionic salt, can dissolve in water at 23°C.
The solubility of most ionic compounds like KBr increases with temperature. This is because increasing temperature provides more kinetic energy to water molecules, allowing them to break the ionic bonds in the salt more efficiently and surround the dissociated ions (K⁺ and Br⁻), resulting in increased solubility.
The question provides a temperature range (20°C to 30°C) and asks us to assume the solubility increases linearly across that range. A linear increase means the solubility change is consistent for each degree Celsius. From solubility data, we know that KBr has a solubility of 65 g per 100 g of water at 20°C and 74 g per 100 g of water at 30°C. The total increase over 10 degrees is 9 g, so the increase per degree is 0.9 g.
Since 23°C is 3°C above 20°C, we multiply the 0.9 g/°C by 3 to get an increase of 2.7 g. Adding this to the base solubility at 20°C (65 g), we get a final solubility of 67.7 g at 23°C.
This method is useful when exact solubility data isn’t available for the target temperature but values are known at two reference points. Linear interpolation is an estimation technique that assumes a straight-line relationship between two known values—a reasonable assumption over small temperature intervals for many salts.