Help! Can someone simplify the steps involved in this please: Given the followin

Question

Help! Can someone simplify the steps involved in this please:

Given the following information: • Concentrated perchloric acid (HClO4) stock = 70% (w/w) • % weight/weight (w/w) will mean number of ‘mass’ parts in 100 ‘mass’ parts • Density of concentrated perchloric acid (70% (w/w)) stock = 1.664 g/mL • Molecular weight: HClO4 = 100.46

a) You are asked to prepare a 300 mL solution of 250 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 200 mL of a pH 2.5 perchloric acid solution?

Explanation / Answer

The density of concentrated perchloric acid (70%, w/w) is 1.664 gm/mL. Therefore, weight of 1 L (=1000 mL) of concentrated perchloric acid is (1.664 gm/1 mL)*(1000 mL) = 1664 gm.

Now, concentrated perchloric acid is 70% (w/w), meaning that 100 gm of concentrated acid contains 70 gm of perchloric acid.

Therefore, 1664 gm of concentrated acid will contains (70 gm/100 gm)*(1664 gm) = 1164.8 gm perchloric acid.

Thus, 1 L of acid solution contains 1164.8 gm perchloric acid.

Molar mass of perchloric acid is 100.46 gm/mol)

Moles of perchloric acid present = (1164.8 gm)*(1 mol/100.46 gm) = 11.594 mole 11.6 mole.

Molarity of stock concentrated perchloric acid = (11.6 mole)/(1 L) = 11.6 M

(a) We want to make 300 mL of 250 mM perchloric acid solution. We know that 1 M solution = 1000 mM solution. Let the volume of stock perchloric acid required be V1 mL.

Using the dilution equation,

(300 mL)*(250 mM)*(1 M/1000 mM) = V1*(11.6 M)

=====> 75 mL = 11.6*V1

=====> V1 = 75/11.6 mL = 6.465 mL 6.5 mL

So, to prepare 300 mL of 250 mM perchloric acid from the supplied stock, we need to add 6.5 mL of the concentrated perchloric acid solution to a 300 mL volumetric flask and make up with distilled water upto the mark. The volume of water to be added = (300 – 6.5) mL = 293.5 mL (ans).

(b) The pH of the solution is 2.5; we know that pH = -log10[H+]

Hence, [H+] = 10-pH = 10-2.5 = 3.16*10-3

The concentration of proton is 3.16*10-3 M

Now, perchloric acid is a very strong acid that undergoes dissociation as

HClO4 (aq) -----------> H+ (aq) + ClO4- (aq)

The molar ratio of HClO4:H+ is 1:1; also, since perchloric acid is a strong acid, the concentration of HClO4 = 3.16*10-3 M.

So, we want to make 200 mL of 3.16*10-3 M acid solution from the stock solution. Let V mL of the the stock solution be required. By using the dilution equation,

(200 mL)*(3.16*10-3 M) = V*(11.6 M)

====> 0.632 mL = 11.6*V

====> V = 0.632/11.6 mL = 0.05448 mL

We are asked to report the volume the L; we can easily do so by noting that 1 mL = 1000 L.

Therefore, volume of the stock solution required = (0.05448 mL)*(1000 L/1 mL) = 54.48 L 54.5 L (ans).

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