To prepare 0.02M NaOH from 1M NaOH solution, you will need to dilute the 1M solution. Use the formula: C1V1 = C2V2, where C1 is the concentration of the stock solution (1M), V1 is the volume of the stock solution you will use, C2 is the desired concentration (0.02M), and V2 is the final volume of the diluted solution. Calculate the volume of 1M NaOH solution (V1) needed to make the desired 0.02M concentration and dilute it with water to reach the desired volume (V2).
To prepare a 1N NaOH solution, you would need to dissolve 40 grams of NaOH in water to make 1 liter of solution. This amount is used because 1N solution means 1 mole of NaOH per liter of solution, and the molar mass of NaOH is 40 g/mol, so 40 grams of NaOH is needed to have 1 mole in 1 liter of solution.
To prepare 10 ml of 1M NaOH solution, you would need to dissolve 0.4 grams of NaOH (molar mass of NaOH is 40 g/mol) in enough distilled water to make a total volume of 10 ml. First, measure out 0.4 grams of NaOH using a balance. Then, add this solid NaOH to a small beaker and add distilled water while stirring until the total volume reaches 10 ml. Make sure to handle NaOH with care as it is a caustic substance.
To prepare 1M Tris-HCl from a 10mM solution, you would need to dilute the 10mM solution by a factor of 100. This means you would mix 1 part of the 10mM solution with 99 parts of water to achieve a final concentration of 1M Tris-HCl.
Mixing equal quantities of 1M HCl and 1M NaOH solutions will give a neutral solution because they will react to form water and a salt (NaCl).
As many as you like, depending on what you want the concentration to be. (Solubility will eventually become a factor, but sodium hydroxide is pretty darn soluble.)
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
To prepare a 1N NaOH solution, you would need to dissolve 40 grams of NaOH in water to make 1 liter of solution. This amount is used because 1N solution means 1 mole of NaOH per liter of solution, and the molar mass of NaOH is 40 g/mol, so 40 grams of NaOH is needed to have 1 mole in 1 liter of solution.
To prepare 10 ml of 1M NaOH solution, you would need to dissolve 0.4 grams of NaOH (molar mass of NaOH is 40 g/mol) in enough distilled water to make a total volume of 10 ml. First, measure out 0.4 grams of NaOH using a balance. Then, add this solid NaOH to a small beaker and add distilled water while stirring until the total volume reaches 10 ml. Make sure to handle NaOH with care as it is a caustic substance.
The molarity of a NaOH solution is determined by the concentration of NaOH in moles per liter of solution. It is calculated by dividing the moles of NaOH by the volume of solution in liters. For example, a 0.1 M NaOH solution would contain 0.1 moles of NaOH per liter of solution.
To prepare 1M Tris-HCl from a 10mM solution, you would need to dilute the 10mM solution by a factor of 100. This means you would mix 1 part of the 10mM solution with 99 parts of water to achieve a final concentration of 1M Tris-HCl.
Mixing equal quantities of 1M HCl and 1M NaOH solutions will give a neutral solution because they will react to form water and a salt (NaCl).
0.1 M sodium hydroxide solution contains 0.1 moles of sodium hydroxide per liter of solution. This corresponds to 0.1 moles of NaOH per 40 g (1 mole) of NaOH, resulting in 4 g of NaOH in 1 liter of 0.1 M NaOH solution.
As many as you like, depending on what you want the concentration to be. (Solubility will eventually become a factor, but sodium hydroxide is pretty darn soluble.)
The conductivity of 1M NaOH solution is relatively high due to the presence of free ions. Sodium hydroxide dissociates completely in water to form Na+ and OH- ions, which are responsible for conducting electricity.
You would need to know the concentration of the unknown HCl solution in order to determine the volume of 1M NaOH needed to neutralize it using the equation: M1V1 = M2V2. Without the concentration of HCl, it is not possible to calculate the volume of NaOH required for neutralization.
As a thought experiment - suppose you mixed one liter of 1M HCl with 1 liter of 1M NaOH. The resulting solution (neglecting any density changes associated with mixing) would contain 2 liters with 1 mole of Cl- and 1 mole of Na+ with the balance being water (the H+ from the HCl and the OH- from the NaOH would just become part of the water). This would give you a 0.5 M solution of NaCl.
To prepare 1M HCl solution from 35% HCl solution, you would need to dilute the 35% HCl with water. Use the formula C1V1 = C2V2, where C1 is the initial concentration, V1 is the volume of the initial solution, C2 is the final concentration (1M), and V2 is the final volume (1 liter in this case). Calculate the volume of 35% HCl needed to achieve a 1M solution, then add water to make up the total volume to 1 liter.