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How would you prepare 250 mL of 0.125 M HCl from concentrated HCl (aq) that is 38.0% by mass with a density of 1.19 g/mLnote: find moles in 250 mL of 0.125 HCl

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Answer

250 mL of 0.125 M HCl can be prepared from concentrated HCl (aq) that is 38.0% by mass with a density of 1.19 g/mL by adding 2.49 mL of HCL(38%) into a small quantity of water, mix to disperse; then dilute with solvent water up to but not to exceed the total needed volume of solution (in this case 250 mL).

Explanation

The problem can be solve in four steps:

Step 1: Calculate the number of moles in 38.0% by mass of HCl.

[tex]\begin{gathered} \text{HCl (38.0}\%)\text{ }=(\frac{38g\text{ HCl}}{100\text{ g solution}})=(\frac{x\text{ moles HCl}}{1.0\text{ Liter Solution}}) \\ \\ \Rightarrow Moles\text{ of HCl in 38 grams }=\frac{38g\text{ HCl}}{36\text{ g/mol}}=1.06\text{ mole HCl} \end{gathered}[/tex]

Step 2: The volume of the 38.0% by mass HCl solution.

Volume of solution in Liters containing 1.06 mole of HCl =

[tex]\frac{Mass}{D\text{ensity}}=\frac{100g}{1.19gmL^{-1}^{}}=84.03\text{ mL HCl(38.0}\%)\text{ }=0.084\text{ L HCl(38.0}\%)[/tex]

Step 3: Calculate the molarity of HCl (38.0%)

[tex]\text{Molarity of HCl (38.0}\%)=\frac{Number\text{ of mole}}{Volume\text{ in liters}}=\frac{1.06\text{ mole HCl}}{0.084\text{ L HCl}}=12.62\text{ M}[/tex]

Step 4: To calculate the volume of HCl (38.0%) required to prepare 250 mL of 0.125 M HCl.

Note: Formula weight of HCl = 36 g/mol

[tex]\begin{gathered} \text{Volume of HCl(38.0}\%)\text{ required }=\frac{(Molarity)(Volume)(Formula\text{ Weight)}}{(\text{Purity)(Specific Gravity)}} \\ \\ \text{Volume of HCl(38.0}\%)\text{ required }=\frac{(0.125\text{ M)})(250\text{ }mL)(36\text{ g/mol)}}{(0.38\text{)}(1.19\text{ g/mL)}} \\ \\ \text{Volume of HCl(38.0}\%)\text{ required }=\frac{(0.125\text{ M)})(0.250\text{ }L)(36\text{ g/mol)}}{(0.38\text{)}1.19\text{ g/mL)}} \\ \\ \text{Volume of HCl(38.0}\%)\text{ required }=\frac{1.125}{0.4522}=2.49\text{ mL} \end{gathered}[/tex]

Therefore, 250 mL of 0.125 M HCl can be prepared from concentrated HCl (aq) that is 38.0% by mass with a density of 1.19 g/mL by adding 2.49 mL of HCL(38%) into a small quantity of water, mix to disperse; then dilute with solvent water up to but not to exceed the total needed volume of solution (in this case 250 mL).

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