Question

Consider the following reaction :

Fe³⁺ (aq) + SCN^- (aq) ---> Fe(SCN)²⁺ (aq)

Starting with 4.00 mL of .200 M Fe³⁺ (aq) in a cuvette, 0.10 mL increments of 0.00100 M SCN^- (aq) will be added. Assume that because [Fe³⁺ (aq)] >> [SCN^- (aq)], the [Fe(SCN)²⁺] concentration can be calculated from the limiting reagent, SCN^-. Calculate [Fe(SCN)²⁺].

Volume .00100 M KSCN mL	[Fe(SCN)2+] (M)
.10
.20
.30
.40
.50
.60
.70
.80
.90
1.00

Answer 1

Solution :-

using the molarity and volume sof the each we calculate the moles of the SCN- and then use them as the moles opf product because mole ratio is 1 :1

then calculate the concnetration of the FeSCN^2+ at total volume in each part

following are the calculations for each part

Moles of Fe^3+ = molarity * volume in liter

                              = 0.200 mol per L * 0.004 L

                              = 0.0008 mol

Moles of SCN- part 1 = 0.001 mol per L * 0.0001 L = 1*10^-7

So moles of FeSCN2+ produced = 1*10^-7 mol

Molarity of this at total volume = 4 ml + 0.1 ml = 4.1 ml = 0.0041 L is

[FeSCN^2+] = 1*10^-6 mol / 0.0041 L = 2.44*10^-5 M

Part 2

Moles of SCN- part 1 = 0.001 mol per L * 0.0002 L = 2*10^-7

So moles of FeSCN2+ produced = 2*10^-7 mol

Molarity of this at total volume = 4 ml + 0.2 ml = 4.2 ml = 0.0042 L is

[FeSCN^2+] = 2*10^-6 mol / 0.0042 L = 4.76*10^-5 M

Part 3

Moles of SCN- part 1 = 0.001 mol per L * 0.0003 L = 3*10^-7

So moles of FeSCN2+ produced = 3*10^-7 mol

Molarity of this at total volume = 4 ml + 0.3 ml = 4.3 ml = 0.0043 L is

[FeSCN^2+] = 3*10^-6 mol / 0.0043 L = 6.977*10^5 M

Part 4

Moles of SCN- part 1 = 0.001 mol per L * 0.0004 L = 4*10^-7

So moles of FeSCN2+ produced = 4*10^-7 mol

Molarity of this at total volume = 4 ml + 0.4 ml = 4.4 ml = 0.0044 L is

[FeSCN^2+] = 4*10^-6 mol / 0.0044 L = 9.091*10^-5 M

Part 5

Moles of SCN- part 1 = 0.001 mol per L * 0.0005 L = 5*10^-7

So moles of FeSCN2+ produced = 5*10^-7 mol

Molarity of this at total volume = 4 ml + 0.5 ml = 4.5 ml = 0.0045 L is

[FeSCN^2+] = 5*10^-7 mol / 0.0045 L = 1.11*10^-4 M

Part 6

Moles of SCN- part 1 = 0.001 mol per L * 0.0006 L = 6*10^-7

So moles of FeSCN2+ produced = 6*10^-7 mol

Molarity of this at total volume = 4 ml + 0.6 ml = 4.6 ml = 0.0046 L is

[FeSCN^2+] = 6*10^-7 mol / 0.0046 L = 1.304*10^-4 M

Part 7

Moles of SCN- part 1 = 0.001 mol per L * 0.0007 L = 7*10^-7

So moles of FeSCN2+ produced = 7*10^-7 mol

Molarity of this at total volume = 4 ml + 0.7 ml = 4.7 ml = 0.0047 L is

[FeSCN^2+] = 7*10^-7 mol / 0.0047 L = 1.489*10^-4 M

Part 8

Moles of SCN- part 1 = 0.001 mol per L * 0.0008 L = 8*10^-7

So moles of FeSCN2+ produced = 8*10^-7 mol

Molarity of this at total volume = 4 ml + 0.8 ml = 4.8 ml = 0.0048 L is

[FeSCN^2+] = 8*10^-7 mol / 0.0048 L = 1.667*10^-4 M

Part 9

Moles of SCN- part 1 = 0.001 mol per L * 0.0009 L = 9*10^-7

So moles of FeSCN2+ produced = 9*10^-7 mol

Molarity of this at total volume = 4 ml + 0.9 ml = 4.9 ml = 0.0049 L is

[FeSCN^2+] = 9*10^-7 mol / 0.0049 L = 1.837*10^-4 M

Part 10

Moles of SCN- part 1 = 0.001 mol per L * 0.001 L = 1*10^-6

So moles of FeSCN2+ produced = 1*10^-6 mol

Molarity of this at total volume = 4 ml + 1.0 ml = 5.0 ml = 0.005 L is

[FeSCN^2+] = 1*10^-6 mol / 0.005 L = 2.00*10^-4 M