Homework Help Question & Answers

Concentration at equilibrium?

The formation constant* of #[M(CN)_6]^(4– #is #2.50 × 10^17#, where M is a generic metal. A 0.150-mole quantity of #M(NO_3)^2 #is added to a liter of 1.220M NaCN solution. What is the concentration of #M^(2+)# ions at equilibrium?

0 0
Add a comment
Answer #1

Warning! Long Answer. The equilibrium concentration of #"M"^"2+"# ions is #8.23 × 10^"-16"color(white)(l)"mol/L"#.

Explanation:

The balanced equation is

#"M"^"2+" + "6CN"^"-" ⇌ "M(CN)"_6^"4+"; K_text(f) = 2.50 × 10^17#

We can set up an ICE table.

#color(white)(mmmmmmml)"M"^"2+" + color(white)(ll)"6CN"^"-" ⇌ "M(CN)"_6^"4+"#
#"I/mol·L"^"-1": color(white)(mll)0.150 color(white)(mm)1.220color(white)(mmmll)0#
#"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmll)"-6"xcolor(white)(mmml)"+"x#
#"E/mol·L"^"-1": color(white)(m)"0.150-"xcolor(white)(m)"1.220-6"xcolor(white)(mml)x#

#K_text(f) = (["M(CN)"_6^"4+"])/(["M"^"2+"]["CN"^"-"]^6) = x/(("0.150-"x)(1.220-6"x)^6) = 2.50 × 10^17#

This result looks rather unpromising. We must solve a seventh-order polynomial.

However, the value of #K_text(f)# is so large that the reaction goes substantially to completion.

Thus, 0.150 mol of #"M"^"2+"# will react with 0.900 mol of #"CN"^"-"# and form 0.150 mol of #"M(CN)"_6^"4+"# (with 0.300 mol of #"CN"^"-"# unused).

We can approach the equilibrium from the other end. Let's assume that the reaction goes to completion and write the equilibrium in reverse.

#"M(CN)"_6^"4+" ⇌"M"^"2+" + "6CN"^"-"#

Then,

#K = 1/K_text(f) = 1/(2.50 × 10^17) = 4.00× 10^"-18"#

Our new ice table becomes

#color(white)(mmmmmm)"M(CN)"_6^"4+" ⇌ "M"^"2+" + "6CN"^"-"#
#"I/mol·L"^"-1": color(white)(mll)0.150 color(white)(llmmm)0color(white)(mml)0.300#
#"C/mol·L"^"-1": color(white)(mm)"-"xcolor(white)(mmmm)"+"xcolor(white)(mm)"+6"x#
#"E/mol·L"^"-1": color(white)(m)"0.150-"xcolor(white)(mmm)xcolor(white)(ml)"0.300+6"x#

#K = (["M"^"2+"]["CN"^"-"]^6)/(["M(CN)"_6^"4+"]) = (x("0.300+6"x)^6)/("0.150-"x)= 4.00 × 10^"-18"#

Check for negligibility

#0.300/(4.00 × 10^"-18") = 7. 5 × 10^16 > 400#. ∴ #6x ≪ 0.300# and #x ≪ 0.150#.

Then,

#(x(0.300)^6)/0.150 = 4.00 × 10^"-18"#

#7.29 × 10^"-4"x = 0.150 × 4.00 × 10^"-18" = 6.00 × 10^"-19"#

#x = (6.00 × 10^"-19")/(7.29 × 10^"-4") = 8.23 × 10^"-16"#

#["M"^"2+"] = 8.23 × 10^"-16"color(white)(l)"mol/L"#

Check:

#K = ((8.23 × 10^"-16")(0.300 + 6 × 8.23 × 10^"-16")^6)/("0.150 -8.23 × 10"^"-16") = ((8.23 × 10^"-16")(0.300)^6)/0.150 = 4.00 × 10^"-18"#

It checks!

answered by: Ernest Z.
Add a comment
Know the answer?
Add Answer to:
Concentration at equilibrium?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coin

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.