how many milligrams of PbI2 can you dissolve in 300 mL of water at 25 degrees celcius
PbI2 ==> Pb^+ + 2I^-
Ksp = (Pb^+2)(I^-)^2
Set up an ICE chart, substitute into Ksp expression, and solve for PbI2 solubility. Post your work if you get stuck.
To determine the maximum amount of PbI2 that can be dissolved in 300 mL of water at 25 degrees Celsius, we need to refer to the solubility data for PbI2.
1. Look up the solubility of PbI2: The solubility of PbI2 is approximately 0.0625 grams per 100 mL of water at 25 degrees Celsius.
This means that for every 100 mL of water, you can dissolve 0.0625 grams of PbI2.
2. Convert the volume: Convert the volume of water from milliliters to liters by dividing 300 mL by 1000:
300 mL ÷ 1000 = 0.3 L
3. Use the solubility data: Multiply the solubility of PbI2 (in grams per 100 mL) by the volume of water (in liters) to find out the maximum amount that can be dissolved.
Maximum amount of PbI2 = Solubility x Volume of water
= 0.0625 g/100 mL x 0.3 L
= 0.0625 g/100 mL x 0.3 L x 1000 mL/1 L
= 0.1875 g
So, you can dissolve a maximum of 0.1875 grams (or 187.5 milligrams) of PbI2 in 300 mL of water at 25 degrees Celsius.