Question

Given the following information:

Concentrated perchloric acid (HClO4) stock = 70% (w/w)

Density of concentrated perchloric acid (70% (w/w)) stock = 1.664 g/mL

Molar mass of HClO4 = 100.46 g/mol

a) You are asked to prepare a 300 mL solution of 400 mM perchloric acid (HClO4). What volumes (in mL) of concentrated HClO4 and water are required to make this solution?

b) What volume (in L) of concentrated perchloric acid (HClO4) stock is required to make 500 mL of a pH 2.7 perchloric acid solution?

Answer 1

First we shall calculate the Molarity of the stock solution.

Since density of the stock solution is = 1.664 g/ml

Therefore; mass of 1 L of stock solution = 1.664 g/ml * 1000 ml = 1664 g

Now 70% (w/w) means there is 70g of HClO₄ in 100 g of stock solution.

mass of the HClO₄ in 1 L stock solution  = (70/100) * 1664 = 1164.8 g

moles of HClO₄ = given mass / molar mass = 1164.8/100.46 = 11.59 mol

therefore, molarity = number of moles / Litre of solution = 11.59 M

a)

now we can use dilution eqaution :

M_dilutionV_dilution = M_stockV_stock

0.4 mol * 0.3 L = 11.59 mol * V_stock

V_stock = 0.0103 L = 10.3 mL

therefore we would need 10.3 mL of HClO₄ stock solution and (300 - 10.3) = 289.7 mL of water to get the desired concentration and volume of HClO₄

b)

pH of the desired solution of HClO₄ = 2.7

Volume of the desired solution = 500 mL

Since HClO₄ is a monoprotic acid the concentration of H⁺ ion will be equal to the concentration of acid solution.

since pH = -log[H⁺]

therefore, [H⁺] = 10^-pH = 10^-2.7 = 0.002 M

Now simply use the dilution equation :

M_dilutionV_dilution = M_stockV_stock

0.002 M * 0.5 L = 11.59 M * V_stock

V_stock = 0.000086 L = 0.086 mL = 86 μL