Supposing you want to get a total volume of 1.00 Litre of 0.01 M HCl (= 0.01 mol/L), you've to take 1/100th of this, being 10.0 mL of the 1M HCLsolution
and add this to about 0.9 L distilled water, mix it and then fill up to exactly 1.00 Litre.
Dilution equation:
V1 * c1 = V2 * c2
Filled in (use same units on each side!):
1.00 (L) * 0.01 (M) = V2 (L) * 1 (M) ==> V2 = 0.01 L = 10.0 mL
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.
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.
using the equation M1V1 = M2V2, V2 = M1V1/M2 = 1 x V/ 0.5 basically, whatever volume of 1M HCl you have, add the same mount of water and you will dilute it to 0.5M.
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.
About 13M. You can assume it is 13M if you don't need an exact concentration (like if you need a ~1M HCl solution for an extraction or whatever) but if you need an exact concentration (for a titration, for example) then you will need to standardize your HCl first.
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.
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.
You could titrate equal volumes of 1M solution of NaOH and 1M solution of HCl to obtain 1M solution of NaCl.
using the equation M1V1 = M2V2, V2 = M1V1/M2 = 1 x V/ 0.5 basically, whatever volume of 1M HCl you have, add the same mount of water and you will dilute it to 0.5M.
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.
About 13M. You can assume it is 13M if you don't need an exact concentration (like if you need a ~1M HCl solution for an extraction or whatever) but if you need an exact concentration (for a titration, for example) then you will need to standardize your HCl first.
.5M
1M HCl means there is 1 mole of HCl in 1 liter of solution. To convert to 1N HCl, you need to consider the equivalent weight of HCl, which is its molecular weight as it is a monoprotic acid. So, in this case, 1M HCl is equivalent to 1N HCl.
Ah, preparing a 0.02 M solution of HCl is a wonderful journey. Simply measure out the correct amount of hydrochloric acid and dilute it with water until you reach the desired concentration. Remember to handle chemicals with care and always wear appropriate safety gear. Happy experimenting, my friend!
To make a 0.1M solution from a 1M HCL solution, you would dilute the 1M HCL with 10 parts of water (or whatever solvent you are using). For example, mix 1 mL of 1M HCL with 9 mL of water to obtain a 0.1M HCL solution.
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).
they both are same as HCl is a monobasic acid.>>>Not exactly. N stands for normal and M stands for mole. Knowing that, read this article to know the difference:http://answers.yahoo.com/question/index?qid=20070625100319AALNjoW