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

Answer:

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.

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 = $\frac{36.0 g}{36.5 g/mol}=0.9863 mol$

Mass of solution ,m= 100 g

Volume of solution = V = ?

Density of the solution ,d= 1.18 g/mL

$V=\frac{d}{M}=\frac{100 g}{1.18 g/mL}=84.75 mL=0.08475 L$

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

Molarity of the solution :

$=\frac{0.9863 mol}{0.8475 L}=11.64 mol/L$

11.64 mol/L is the molarity of concentrated HCl.

2)

$M_1V_1=M_2V_2$ ( Dilution equation)

$M_1= 11.64 M$

$V_1=?$

$M_2=1.6 M$

$V_2=985 mL$

$V_1=\frac{M_2V_2}{M_1}=\frac{1.6 M\times 985 mL}{11.64 M}$

$=135.40 mL$

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

$11.64 M=\frac{n}{1.75 L}$

$n=1.75 L\times 11.64 M=20.37 mol$

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

Then 20.37 moles of HCl will neutralized by ;

$\frac{1}{1}\times 20.37 mol=20.37 mol$ 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.