Saturated solutions of each of the following compounds are made at 20 degree Celsius. Which of the solutions will form a precipitate upon heating?
a. NaCl b. Na2SO4
c. Li2CO3 d. sucrose
Since water heats at 100 degree Celsius here are there solubility (g/100 g of H2O) measurements
for a. 39.2 b. 41.0 c. 0.70 d. 487
Can you explain to me how you got the answer?
Frankly, I don't understand the data table. It's nice to know the solubility but what is the temperature? 100 C perhaps? And what is the saturated solution of each? The answer, I believe, is Na2SO4. The solubility curve for Na2SO4.10H2O decreases above about 30 C.
http://en.wikipedia.org/wiki/Sodium_sulfate
Of course, I'd be happy to explain it to you! When determining if a precipitate will form upon heating, we need to compare the solubility of the compound in its saturated solution at room temperature to its solubility at the higher temperature, which in this case is the boiling point of water.
Let's look at each compound individually:
a. NaCl: The solubility of NaCl at 20°C is 39.2 g/100 g of H2O. This means that 39.2 grams of NaCl can dissolve in 100 grams of water at 20°C. Since the solubility of NaCl is quite high, it will not form a precipitate when heated.
b. Na2SO4: The solubility of Na2SO4 at 20°C is 41.0 g/100 g of H2O. Similar to NaCl, Na2SO4 also has a high solubility and will not form a precipitate upon heating.
c. Li2CO3: The solubility of Li2CO3 at 20°C is 0.70 g/100 g of H2O. This solubility is relatively low, indicating that Li2CO3 is not very soluble in water at room temperature. When the solution is heated, the solubility of Li2CO3 will likely decrease even further, leading to the formation of a precipitate.
d. Sucrose: The solubility of sucrose (C12H22O11) at 20°C is 487 g/100 g of H2O. Like NaCl and Na2SO4, sucrose has a high solubility and will not form a precipitate upon heating.
Therefore, the compound that will form a precipitate upon heating is c. Li2CO3.
To determine which solution will form a precipitate upon heating, we need to compare the solubility of each compound in water at 20 degrees Celsius to its solubility at 100 degrees Celsius. If the solubility decreases significantly with increasing temperature, it indicates that the compound is less soluble when the solution is heated and is likely to form a precipitate.
a. NaCl:
The solubility of NaCl is 39.2 g/100 g of water at 20 degrees Celsius. Since NaCl is a common salt, it remains highly soluble even at higher temperatures. Therefore, it will not form a precipitate upon heating.
b. Na2SO4:
The solubility of Na2SO4 is 41.0 g/100 g of water at 20 degrees Celsius. Similar to NaCl, Na2SO4 is also highly soluble and will not form a precipitate upon heating.
c. Li2CO3:
The solubility of Li2CO3 is 0.70 g/100 g of water at 20 degrees Celsius. Given its relatively low solubility, there is a possibility that Li2CO3 could become less soluble at higher temperatures and form a precipitate upon heating.
d. Sucrose:
Sucrose, commonly known as table sugar, has a solubility of 487 g/100 g of water at 20 degrees Celsius. This high solubility indicates that sucrose remains highly soluble even at higher temperatures and will not form a precipitate upon heating.
In summary, only Li2CO3 has a low solubility that may decrease significantly with increasing temperature, making it likely to form a precipitate when heated.
To determine which of the solutions will form a precipitate upon heating, we need to compare the solubility of each compound at 20°C with their respective solubilities at 100°C.
For compound a (NaCl), the solubility at 20°C is 39.2 g/100 g of water. Since the solubility of NaCl at 100°C is significantly higher than its solubility at 20°C, it means that NaCl would remain dissolved even upon heating, so no precipitate would form.
For compound b (Na2SO4), the solubility at 20°C is 41.0 g/100 g of water. We need to check if the solubility at 100°C is lower than this value. However, if we assume that Na2SO4's solubility decreases with increasing temperature, we can conclude that no precipitate would form upon heating.
For compound c (Li2CO3), the solubility at 20°C is 0.70 g/100 g of water. Since this solubility is relatively low, it is likely that the solubility of Li2CO3 at 100°C is even lower. Thus, if we heat the saturated solution, it is highly probable that Li2CO3 would form a precipitate.
For compound d (sucrose), the solubility at 20°C is 487 g/100 g of water. Sucrose is a carbohydrate and its solubility tends to increase with increasing temperature. Therefore, it is highly unlikely that a precipitate would form if we heat the saturated sucrose solution.
In summary, compound c (Li2CO3) is the only one that will form a precipitate upon heating.