Question

How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce 20.0 L of a solution that has a pH of 1.80?

Answer 1

(1) Calculation of molarity of pH=1.80 solution :

pH = -log[H⁺]

---> [H⁺] = 10 ^-pH

                = 10^-1.80

                = 0.016 M
(2) Calculation of molarity of 36% HCl solution ( stock solution) :
Given density = 1.18g/mL

So for 1000 mL of the solution , the mass of HCl is m = density x volume

                                                                                      = 1.18 g/mL x 1000 mL

                                                                                      = 1180 grams
But the solution is 36% HCl thus the actual weight is = (36/100) x1180 g = 424.8 g

Molar mass of HCl = 1 + 35.5 = 36.5 g/mol

Then number of moles , n = mass/molar mass

                                         = 424.8 g/ 36.5 (g/mol)

                                         = 11.64 mol

So Molarity of the stock solution , M = number of moles / volume in L

                                                         = 11.64 mol / 1.0 L

                                                         = 11.64 M

According to law of dilution   MV = M'V'

Where M = Molarity of stock = 11.64 M

V = Volume of the stock = ?

M' = Molarity of dilute solution = 0.016 M

V' = Volume of the dilute solution = 20.0 L = 20.0x1000 = 20000 mL

Plug the values we get   , V = M'V' / M

                                            = ( 0.016 x 20000 ) / 11.64

                                           = 27.5 mL

Therefore the volume of solution is 27.5 mL