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40 grams, this is the 1M NaOH standard laboratory solution.
This sodium hydroxide solution has a molarity of 0,25.
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.
THE PH VALUE ACIDIC SOLUTION VARIOUS FROM 0-6.9, WHILE THE BASIC SOLUTION VARIOUS FROM 7.1-1.4. THUS ,OUT OF HCL AND NaOH WILL HIGHER PH VALUE
In order to prepare exactly 30 mL of 1M NaOH solution, a volumetric vessel that contains exactly 30 mL when full to a marked level will be needed. By definition, a 1 M solution contains one mole in a liter of volume, and since solutions are homogeneous, 30 mL of such a solution will require (30/1000) mole of sodium hydroxide. The molar mass of NaOH is about 40, corresponding to 1.2 grams of sodium hydroxide, which can be determined by weighing solid sodium hydroxide, dissolving the weighed amount in a volume of water less than the 30 mL capacity of the volumetric vessel, transferring this more concentrated solution into the volumetric vessel, and diluting with pure water until the 30 mL volume is contained in the vessel.
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
40 grams, this is the 1M NaOH standard laboratory solution.
One mole of NaOH is 40g. So 40g dissolved in 1 litre will give a 1M solution.
This sodium hydroxide solution has a molarity of 0,25.
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.
What is the conductivity of 1 molar solution of sodium hydroxide at ambient temperature
THE PH VALUE ACIDIC SOLUTION VARIOUS FROM 0-6.9, WHILE THE BASIC SOLUTION VARIOUS FROM 7.1-1.4. THUS ,OUT OF HCL AND NaOH WILL HIGHER PH VALUE
In order to prepare exactly 30 mL of 1M NaOH solution, a volumetric vessel that contains exactly 30 mL when full to a marked level will be needed. By definition, a 1 M solution contains one mole in a liter of volume, and since solutions are homogeneous, 30 mL of such a solution will require (30/1000) mole of sodium hydroxide. The molar mass of NaOH is about 40, corresponding to 1.2 grams of sodium hydroxide, which can be determined by weighing solid sodium hydroxide, dissolving the weighed amount in a volume of water less than the 30 mL capacity of the volumetric vessel, transferring this more concentrated solution into the volumetric vessel, and diluting with pure water until the 30 mL volume is contained in the vessel.
I think you may have missed a decimal point somewhere. 125M of NaOH would be a solution of sodium hydroxide containing 125 moles per litre. One mole of a compound is the same number of grams as the molecular weight of the molecule. Sodium hydroxide has a molecular weight of 40 ( sodium 23, oxygen 16, and hydrogen 1), so a one molar solution would have forty grams of NaOH per litre. 500ml of a 1M solution would contain 20g. 500ml of a 125M solution would need 2 500g. 1L of a 125M solution would need 5 000g of sodium hydroxide in the litre. The maximum solubility for NaOH in water at 20 degrees is 1110g per litre, so if you tried to dissolve 5 000g in a litre you would be left with 3 890g undissolved. A 1.25M solution would have 1.25 times 40g per litre, which is 50g per litre. 500ml of this solution would have half this amount of NaOH, or 25g.
1 millimolar = 0.001 M NaOH ( a base, remember ) - log(0.001 M NaOH) = 3 14 - 3 = 11 pH ----------
All one to one here. NaOH + HCl --> NaCl + H2O So, getting the moles of either product tell you how many moles H2O produced. Molarity = moles of solute/Liters of solution moles of solute = Liters of solution * Molarity ( 30.00 mL = 0.03 liters ) Moles H2O = 0.03 Liters * 1 M 0.030 moles H2O produced =================
1N HCl is also 1M HCl because it is mono-protic. Therefore 36.5 g of HCl is required per liter or 3.65%. Simply take 100 g of 37% HCl and make up to the 1 liter mark on the volumetric flask. Check the value by titration against 1M NaOH. It should be perfect. If very slightly strong dilute very slightly (calculate) with water and re-standardize.