Calculate the amount (in grams) of Tris Base needed to make 1L of 0.01M Tris buffer?
Tris Base=121.4 g/mol
M x V = 0.01 mole/liter x 1 liter = 0.01 mole
mass (g) = mole x molar mass = 0.01 mole x 121.4 grams/mole = 1.214 grams = 1 gram (corrected to 1 significant digit)
M x V = 0.01 mole/liter x 1 liter = 0.01 mole
mass (g) = mole x molar mass = 0.01 mole x 121.4 grams/mole = 1.214 grams = 1 gram (corrected to 1 significant digit)
The solution is not really a buffer. There is no pH mentioned in the problem.
To calculate the amount of Tris Base needed to make 1L of 0.01M Tris buffer, you can use the formula:
Mass (g) = Molarity (mol/L) x Volume (L) x Molecular Weight (g/mol)
Given:
Molarity = 0.01M
Volume = 1L
Molecular Weight of Tris Base = 121.4 g/mol
Substituting these values into the formula:
Mass (g) = 0.01 mol/L x 1 L x 121.4 g/mol
Mass (g) = 1.214 g
Therefore, you would need 1.214 grams of Tris Base to make 1L of 0.01M Tris buffer.
To calculate the amount of Tris Base needed to make 1L of 0.01M Tris buffer, you'll need the formula weight of Tris Base and the desired final molarity.
First, convert the desired final molarity (0.01M) to moles per liter (mol/L). Since Molarity (M) is defined as moles of solute per liter of solution, you can simply use the given value of 0.01M.
Next, use the formula weight of Tris Base (121.4 g/mol) to convert the desired moles per liter to grams. The formula weight represents the mass of one mole of the compound.
To calculate the grams of Tris Base needed, use the formula:
grams of Tris Base = (moles of Tris Base) x (formula weight of Tris Base)
Substituting the given values:
grams of Tris Base = (0.01 mol/L) x (121.4 g/mol)
grams of Tris Base = 1.214 g
Therefore, you need 1.214 grams of Tris Base to make 1L of 0.01M Tris buffer.