Question

According to a label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density of 1.18 g/mL.

1. What is the molarity of concentrated HCl.
2. What volume of it would you need to prepare 985 mL of 1.6 M HCl?
3. What mass of sodium bicarbonate would be needed to neutralize the spill if a bottle containing 1.75 L of concentrated HCl dropped on a lab floor and broke open?

Answer 1

1) 11.64 mol/L is the molarity of concentrated HCl.

2) 135.40 mL of volume of 11.64 M will need to prepare 985 mL of 1.6 M HCl.

3) 1,711.08 grams of sodium bicarbonate would be needed to neutralize the spill HCl solution.

Step-by-step explanation:

HCl solution with 36.0% HCl by mass, menas that in 100 g of solution 36.0 gram of HCl is present.

Mass of HCl= 36.0 g

Moles of HCl =
`(36.0 g)/(36.5 g/mol)=0.9863 mol`

Mass of solution ,m= 100 g

Volume of solution = V = ?

Density of the solution ,d= 1.18 g/mL

`V=(d)/(M)=(100 g)/(1.18 g/mL)=84.75 mL=0.08475 L`

`Molarity=\frac{Moles }{\text{Volume of solution (L)}}`

Molarity of the solution :

`=(0.9863 mol)/(0.8475 L)=11.64 mol/L`

11.64 mol/L is the molarity of concentrated HCl.

2)

`M_1V_1=M_2V_2` ( Dilution equation)

`M_1= 11.64 M`

`V_1=?`

`M_2=1.6 M`

`V_2=985 mL`

`V_1=(M_2V_2)/(M_1)=(1.6 M* 985 mL)/(11.64 M)`

`=135.40 mL`

135.40 mL of volume of 11.64 M will need to prepare 985 mL of 1.6 M HCl.

3.

`NaHCO_3+HCl\rightarrow NaCl+H_2O+CO_2`

Concentration of HCl solution = 11.64 M

Volume of the HCl solution = 1.75 L

Moles of HCl in 1.75 L solution = n

`11.64 M=(n)/(1.75 L)`

`n=1.75 L* 11.64 M=20.37 mol`

According to reaction 1 mole of HCl neutralized by 1 mole of sodium carbonate.

Then 20.37 moles of HCl will neutralized by ;

`(1)/(1)* 20.37 mol=20.37 mol` of sodium carbonate

Mass of 20.37 moles of sodium carbonate :

= 20.37 mol × 84g/mol = 1,711.08 g

1,711.08 grams of sodium bicarbonate would be needed to neutralize the spill HCl solution.