I assume you want to prepare HCl 0.5 M from HCl 37% (that is, concentrated hydrochloric acid). This is a dilution problem: HCl 37% can be assumed to be about 12 M, so it has to be diluted down to 0.5 M. Use this equation:
M1·V1 = M2·V2
M1 = initial concentration (here, 12 M)
V1 = amount of HCl 12 M you need to pick up and, subsequently, dilute
M2 = final concentration (here, 0.5 M)
V2 = amount of HCl 0.5 M you want to make
Solve for V1, you will have the amount (either in mL or L) of HCl 12 M you must use to make up the final solution till the desired volume.
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I assume you want to prepare HCl 0.5 M from HCl 37% (that is, concentrated hydrochloric acid). This is a dilution problem: HCl 37% can be assumed to be about 12 M, so it has to be diluted down to 0.5 M. Use this equation:
M1·V1 = M2·V2
M1 = initial concentration (here, 12 M)
V1 = amount of HCl 12 M you need to pick up and, subsequently, dilute
M2 = final concentration (here, 0.5 M)
V2 = amount of HCl 0.5 M you want to make
Solve for V1, you will have the amount (either in mL or L) of HCl 12 M you must use to make up the final solution till the desired volume.
Preparation and standardisation of 0.5 M Methanolic Hydrochloric Acid using anhydrous Sodium Carbonate and Methyl Red solution as indicator used in analysis of pharmaceutical ingredients.
Methanolic Hydrochloric Acid Solution Preparation
- Take 40 ml of water in a 1000 ml volumetric flask.
- Slowly add 43 ml of hydrochloric acid.
- Cool and add methanol to volume.
- Standardize the solution in the following manner.
Methanolic Hydrochloric Acid Solution Standardization
- Weigh accurately about 800 mg of anhydrous sodium carbonate, previously heated at about 270°C for 1 hour.
- Dissolve it in 100 ml of water and add 0.1 ml of methyl red solution.
- Add the acid slowly from a burette, with constant stirring, until the solution becomes faintly pink.
- Heat the solution to boiling, cool and continue the titration.
- Heat again to boiling and titrate further as necessary until the faint pink color is no longer affected by continued boiling.
- 1 ml of 0.5 M hydrochloric acid is equivalent to 0.02650 g of Na2CO3.
- Calculate the molarity of solution by the following formula:
Na2CO3 in mg
M = --------------------------
Chemistry Quick Review of Solution Preparation
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Here's a quick overview of how to prepare a solution when the final concentration is expressed as M or molarity.
You prepare a solution by dissolving a known mass of solute (often a solid) into a specific amount of a solvent. One of the most common ways to express the concentration of the solution is M or molarity, which is moles of solute per liter of solution.
Example of How to Prepare a Solution
Prepare 1 liter of 1.00 M NaCl solution.
First, calculate the molar mass of NaCl which is the mass of a mole of Na plus the mass of a mole of Cl or 22.99 + 35.45 = 58.44 g/mol
- Weigh out 58.44 g NaCl.
- Place the NaCl in a 1-liter volumetric flask.
- Add a small volume of distilled, deionized water to dissolve the salt.
- Fill the flask to the 1 L line.
If a different molarity is required, then multiply that number times the molar mass of NaCl. For example, if you wanted a 0.5 M solution, you would use 0.5 x 58.44 g/mol of NaCl in 1 L of solution or 29.22 g of NaCl.
Important Points to Remember
- Molarity is expressed in terms of liter of solution, not liters of solvent. To prepare a solution, the flask is filled to the mark. In other words, it is incorrect to a 1 liter of water to a mass of sample to prepare a molar solution.
- Sometimes it's necessary to adjust the pH of a solution. To do this, add enough water to dissolve the solute. Then add an acid or base solution dropwise (usually a hydrochloric acid or HCl solution for acid or sodium hydroxide or NaOH solution for a base) to reach the desired pH. Then add more water to reach the mark on the glassware. Adding more water won't change the pH value.
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What volume of $0.5\ \mathrm M$ HCl solution is needed to prepare a $500\ \mathrm{ml}$ $10\ \%$ concentration solution, whose density is $1.05\ \mathrm{g/cm^3}$.
I've started off with finding the mass of the final solution, which would be $500\ \mathrm{cm^3} \times 1.05\ \mathrm{g/cm^3} = 525\ \mathrm g$. Next I find the mass of the HCl in it, $52.5\ \mathrm g$. After that, the moles of HCl, approx. $1.4384\ \mathrm{mol}$.
But after that I'm not sure how to proceed. I've tried dividing the molarity by the moles to get the volume, but that didn't work. I think I might have to make a system of equations but I'm bad at math so if that's how it's supposed to be done, I'd like for someone to show me how.
asked May 1, 2019 at 7:50
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As you have calculated, 500 ml of a 10 % HCl solution contain about 1.44 mol HCl. Thus the concentration is about 2.9 mol/l. Therefore, you cannot prepare this solution by diluting a solution with c = 0.5 mol/l HCl.
answered May 1, 2019 at 8:01
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