Question

Suppose a student mixes 4.00 mL of 2.00 # 10-3 M Fe(NO3 ) 3 with 5.00 mL of 2.00 # 10-3 M KSCN and 1.00 mL of water. The student then determines the [FeNCS2+] at equilibrium to be 8.75 # 10-5 M. Find the equilibrium constant for the following reaction. Show all your calculations for each step. Fe3+ (aq) + SCN- (aq) FeNCS2+ (aq) Step 1. Calculate the initial number of moles of Fe3+ and SCN- (use Equation 12). moles of Fe3+ _____________ moles of SCN- ______________ Step 2. How many moles of FeNCS2+ are present at equilibrium? What is the volume of the equilibrium mixture? ________________ mL moles of FeNCS2+ ____________ How many moles of Fe3+ and SCN- are consumed to produce the FeNCS2+? moles of Fe3+ _____________ moles of SCN- ______________

Answer 1

$\mathrm{Fe} 3+(\mathrm{ag})+\mathrm{SCN}-(\mathrm{aq})$ <----> $\mathrm{FeSCN} 2+(\mathrm{aq})$

Explanation:

Use the concentration and volume values they've given you to find the moles of Fe(NO3)3 and KSCN that were initially present (before they were mixed)

moles = concentration in M x volume in L

n(Fe(NO3)3) = $\left(2.00 \times 10^{\wedge}-3\right) \times(5 / 1000)=0.00001 \mathrm{mol}$ $=10^{\wedge}-5 \text { mol }$

n(KSCN) = $\left(2.00 \times 10^{\wedge}-3\right) \times(4 / 1000)=0.000008 \mathrm{mol}$ $=8 \times 10^{\wedge}-6 \text { mol }$

n(Fe(NO3)3) = n(Fe3+) = $10^{\wedge}-5 \text { mol }$

$n(KSCN) = n(SCN-) =$ $8 \times 10^{\wedge}-6 \text { mol }$

so n(Fe3+) initially present = $10^{\wedge}-5 \text { mol }$

n(SCN-) initially present = $8 \times 10^{\wedge}-6 \text { mol }$