the solubility of silver sulfate in water is 0.223%(w/v) at 35 degrees celsius.calculate the solubility product of this salt at this temp.
I got 1.46 X 10^-6 M
is this right please
thanks Andy
To calculate the solubility product of silver sulfate (Ag2SO4), you need to convert the given solubility from a percentage to molarity at 35 degrees Celsius.
First, convert the solubility of 0.223% (w/v) to grams of Ag2SO4 dissolved in 100 mL of water:
0.223% (w/v) = 0.223 g/100 mL
Next, use the molar mass of Ag2SO4 to convert grams to moles:
Ag2SO4 molar mass = (2 x atomic mass of Ag) + atomic mass of S + (4 x atomic mass of O)
= (2 x 107.87 g/mol) + 32.06 g/mol + (4 x 16.00 g/mol)
= 311.80 g/mol
Moles of Ag2SO4 = (0.223 g / 100 mL) / (311.80 g/mol)
= 0.000715 mol / 0.1 L
= 0.00715 M
Now, calculate the concentration of Ag+ ions and SO4^2- ions:
Ag2SO4 dissociates into 2 Ag+ ions and 1 SO4^2- ion.
Therefore, concentration of Ag+ ions = 2 x 0.00715 M
= 0.0143 M
And concentration of SO4^2- ions = 0.00715 M
Finally, calculate the solubility product (Ksp) by taking the product of the concentrations of Ag+ and SO4^2- ions:
Ksp = [Ag+]^2 * [SO4^2-]
= (0.0143)^2 * 0.00715
= 0.000289 M^3
So the solubility product of silver sulfate at 35 degrees Celsius is approximately 2.89 x 10^(-4) M^3.
Based on your calculation of 1.46 x 10^(-6) M, it seems like you made a calculation error. Double-check the steps and calculations to ensure accuracy.