Answer to Question #135733 in General Chemistry for Aisha

Question

Student A prepared a 0.038M of the stock sodium salicylate solution using a pipette, transferred 5mL of the 0.038M sodium salicylate stock solution into a 100mL volumetric flask labelled as sample 1. The solution was diluted to the 100ml mark with acidified 0.02M iron (III) chloride solution. What is the concentration of Iron (III) salicylate complex (in M) in the sample 1 flask. Enter the answer in 4 decimal places

Answer 1

STEP I: BALANCED EQUATION

3C₇H₅NaO_3(aq) + FeCl_3(aq) $\to$ Fe(C₇H₅O₃)_{3 (aq)} + 3NaCl_(aq)

From the equation, 3 moles of C₇H₅NaO3 is required to react with only 1 mole of FeCl3 to produce 1 mole of the Iron (III) salicylate complex.

STEP II : CALCULATE MOLES OF C₇H₅NaO₃

From the concentration C₇H₅NaO₃ ;

0.038 moles were in 1000mL

how many moles would be in 5mL?

$=\dfrac{0.038molx5mL}{1000mL}=1.9x10^-4moles$

STEP III : CALCULATE MOLES OF FeCl₃ REQUIRED

From mole ratio of 1:3

Moles of FeCl₃ required$=\dfrac{1}{3}x1.9x10^-4mol =6.33x10^-5moles$

But the moles of FeCl₃ already available in the flask are?

From the concentration and volume of FeCl₃ we know that; (95 mL of FeCl₃ was added)

0.02 moles are in 1000mL

How many moles therefore would be in the 95mL?

$=\dfrac{0.02molx95mL}{1000mL}=1.9x10^-3moles$

That means that FeCl₃ was in excess and the mass of the final moles of the product will depend on the moles of the sodium salicylate solution (the limiting reagent).

STEP IV : CALCULATE MOLES OF THE PRODUCT

Moles of Iron (III) salicylate complex from the 1:1 mole ratio is therefore:

$=\dfrac{1}{1}xmoles ofC7H5NaO3$

$=\dfrac{1}{1}x1.9x10^-4$

= 1.9 x 10^-4 moles.

STEP V : CONVERT THESE MOLES TO MOLARITY

But these moles (1.9x10^-4) were in 100mL of solution?

How many moles would be in 1000mL

$=\dfrac{(1.9x10^-4mol)x1000mL}{100mL}=1.9x10^-3M$

The concentration of the Iron (III) salicylate complex is therefore; 1.9x10^-3M