Question

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

Answer 1

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).

Step-by-step explanation

The problem can be solve in four steps:

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

`\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}`

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

Volume of solution in Liters containing 1.06 mole of HCl =

`\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}\%)`

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

`\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}`

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

`\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 }=(1.125)/(0.4522)=2.49\text{ mL} \end{gathered}`

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).