0.1 M HCl =============
Confirmed.
standarization of 0.1 n hcl against of borax
1st Get the balanced equation NaOH + HCl -> NaCl + H2O Find the number of moles in HCl; n = cv n = 0.46x0.61 n = 0.2806 moles the number of moles of HCl and NaOH is the same so 0.2806moles will be needed
Add 5 litres water.
0.1 M HCl =============
Confirmed.
standarization of 0.1 n hcl against of borax
1st Get the balanced equation NaOH + HCl -> NaCl + H2O Find the number of moles in HCl; n = cv n = 0.46x0.61 n = 0.2806 moles the number of moles of HCl and NaOH is the same so 0.2806moles will be needed
Add 5 litres water.
Mix 125 mL 0,1 N HCl with 125 mL water.
g HCl solution = 2500 mL of HCl * 1 liter/1000 mL * 1190 g/L = 2975 g 37% solution (37 g HCl/100 grams of solution) gives you the grams of HCl: g HCl = 0.37 * 2975 g = 1100.8 g HCl Moles HCl = 1100.8/(36.46 g/mole) = 30.2 moles Therefore the molarity, which equals the normality in this case = 30.2 moles/2.5 L = 12.07 M = 12.07 N If you want to make 100 mL of a 0.1 N solution, Volume of HCl solution needed = (0.1 N * 100 mL) /12.07 N = 0.83 mL Take 0.83 mL of the 37% HCl, and dilute it with water to 100 mL.
N-(3-dimethylaminopropyl)-n'-ethylcarbodiimide hydrochloride
520 ml of HCl in 480 ml of water=1000ml = 5 N
Mix approx. 12 mL of HCl 30 % in 1 L water.
I'm not sure about the 37 thing but here is from NIOSH method in how to prepare 6N HCL -pipette 25.64 mL of 11.7 N (37% HCL fuming) to 50 mL volumetric flask and top up with distilled water GOOD LUCK Always add acid to water.
add 5 ml of 37% HCl to 495 ml Water. This is 0.12 N ;)