How do you prepare 0.5M of an HCl solution?

There is a very useful formula for this type of questions:

v1 * c1 = v2 * c2

v1 and v2 are the volumina of your respective solutions expressed in litres, c1 and c2 their respective concentrations in moles per litre.

Therefore, we plug in the values you have given:

v1 = 1L, c1 = 0.25M, c2 = 1.0M

This results in

1L * 0.25M = v2 * 1.0M

This gives v2 = 0.25L

Therefore, you will need to take 0.25L of the 1M HCl solution, and dilute this 0.25L to a final volume of 1L.

I hope I could help you, and have a nice day!

Well, the concentration of hydrochloric acid is TWICE that of the sodium hydroxide solution…and we know that [math]HCl(aq)[/math] and [math]NaOH(aq)[/math] react in a 1:1 molar ratio according to the equation…

[math]HCl(aq) + NaOH(aq) ightarrow NaCl(aq) + H_{2}O[/math]

And so if we got [math]1•L[/math] of [math]NaOH(aq)[/math] at [math]0.25•mol•L^{-1}[/math] concentration, we require a [math]frac{1}{2}[/math] litre volume of [math]0.50•mol•L^{-1}..[/math]

As always, [math] ext{Molarity}=dfrac{ ext{Moles of solute}}{ ext{Volume of solution}}[/math]

And thus, given TWO of the THREE quantities, we can access the third by division or multiplication…

What's 36% w/w or w/v and what volume do you want to make of 0.5% HCl?

Such a solution is usually prepared by diluting the HCl from a stock solution.

Purchased concentrated HCl is usually 12.1 M.

The question does not state how much acid to prepare so I will answer how to make one liter of 0.5 M HCl.

Using the dilution formula M1V1=M2V2

12.1 M x V1. = 0.5 M x 1.0 L

V1 = 0.041 L which is 41 ml.

I would get a 1 liter volumetric flask and half fill it with water. Carefully in a fume hood add the 41 ml of con HCl then fill the remainder of the volumetric flask to the mark at 1 Liter. Shake it thoroughly and be sure to label it!

You don't. When hydrogen chloride (the gas) is reacted with water, the highest concentration that can be achieved is 12.4 mol/l (or 38 mass%). So it is impossible to prepare 25N HCl solution.

You will need some concentrated HCl, say 6 molar. The question becomes: How many mL of 6 M HCl is required to prepare 150 mL of 0.25 M HCl?

Use the equation: mLa x Ma = mLb x Mb

150 x 0.25 = n x 6;

n = (150 x 0.25)/6 = 6.3 mL.

Carefully mix 6.3 mL of 6 M HCl with 143.7 mL of water. If the preparation is performed accurate measurements, you will have 150 mL of 0.25 M HCl.

To prepare 1 L of 0.5 N HCl solution, you have to take 0.5 gram equivalent of pure HCl.

1 gram equivalent of HCl = 1 mole of HCl.

So, 0.5 gram equivalent HCl = 0.5 mole HCl = (0.5 x 36.5) gram HCl = 18.25 g HCl.

Take 1 L volumetric flask. Pour 18.5 g HCl in it. Add distilled water with shaking, make up the volume upto 1L mark. Your desired solution is ready.

Hope, this helps.

You need to decide how much 0.05 M HCl you want to prepare. Let’s say you need 500 mL.

You need a source of HCl of known concentration. Let’s say you obtain a solution of certified 5.0 M HCl.

The equation for problems like this is

M1 x V1 = M2 x V2, in this case:

5.0 M HCl x # mL = 0.05 M x 500 mL

This is to say: How many mL of 5.0 M acid is required to prepare 500 mL of 0.05 M acid ?

# mL = (0.05 M x 500 mL)/ 5.0 = 5.0 mL

5.0 mL of 5.0 M acid is required to prepare 500 mL of 0.05 M acid.

In practice, you would obtain a 500 mL volumetric flask. Add an amount of water, say 300 mL. Carefully add 5.0 mL of 5.0 M acid. Add water to bring the total volume to 500 mL.