Question

According to the 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. What is the molarity of concentrated HCl? What volume of it would you need to prepare 759 mL of 2.00 M HCl? 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

a)

Molarity = mol of solute / Liter solution

assume a basis of 100g, 36 is HCl

mol = mass/MW = 36/36.5 = 0.986

V = mass/D= 100/1.18 = 84.745 mL = 84.745*!0^-3 L

[HCl] = 0.986/(84.745*10^-3) = 11.63 M

b)

find V for

We need to apply dilution law, which is based on the mass conservation principle

initial mass = final mass

this apply for moles as weel ( if there is no reaction, which is the case )

mol of A initially = mol of A finally

or, for this case

moles of A in stock = moles of A in diluted solution

Recall that

mol of A = Molarity of A * Volume of A

then

moles of A in stock = moles of A in diluted solution

Molarity of A in stock * Volume of A in stock = Molarity of A in diluted solution* Volume of A in diluted solution

Now, substitute known data

Mstock

11.63 *Vstock = 2*759

Vstock = 2*759/11.63 = 130.5 mL required

Q3.

NaHCO3 required to neturalize 1.75 L of concentrated HCl

mol of HCl = MV = 11.63 *1.75 = 20.3525 mol of HCl

mol of NaHCO3 = mol of HCl = 20.3525

mol of NaHCO3 =20.3525

mass = mol*MW = 20.3525*84.007 = 1709.752 g