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53% D Sat Nov 18 History Bookmarks Develop Window Helo saplinglearning.com Apple Cloud Yahoo Bing Google Wikipedia Google Yahoo University of California- Merced- CHEM 10 (INCLUSIVE ACCESS)-Fal17-ROBLE Jump to Fen Resources Gradebook O Assignn 11/20/2017 11:55 PM 0/100 -Prirr @caoul tr Perode Table Available Due Date: Points Pos Grade Cate Description Policies: Question 9 of 12 Map Sapling Learming The formation constant. of NCLis 2.50-10", whoro M is a generic metal A 0.160-mole quantty of MINOsh is added to a litor of 1.190 M NaCN solution What is the concentration of M ions at equilbrium? You can ches You can view give up on a You can keep until you get You lose 5% answer in you attempt at tha o eTextbook O Help With T O Web Help & O Technical su Prsvious Give Up & View Solution Check Ancaer Next xt 80

Explanation / Answer

M+2 + 6CN- --->(CN6)]^4-

Mol of M(NO3)2 = 0.160

Calculate the moles of NaCN:

1liter of 1.190 M NaCN solution (given)

n = CV = 1.190 mol/L x 1L = 1.190 mol

Moles of CN- = 1.190

Since the Kf is extremely large assume that all of the M+2 in M(NO3)2 reacts completely with the CN- ion

1 mol of M2+ reacts with 6 mol of CN- to form 1 mol of [M(CN)6]^-4

0.160 moles M+2 reacts with ( 6 x 0.160 ) 0.960 moles of CN- to form 0.160 moles [M(CN)6]^-4

0.160 moles M+2 reacts with 0.960 moles of CN- to form 0.160 moles [M(CN)6]^-4

Moles of CN- consumed = 0.960 moles

Moles of CN- remaining: 1.190 – 0.960 = 0.230 moles

Kf = [M(CN)6]^-4/ [M+2][CN-]^6

2.50 x 10^17 = 0.160/[M+2](0.230)^6

[M+2] = 0.160 / (0.230)^6 *(2.5 x 10^17)

[M+2] = 0.160 /1.480 x 10^-4*(2.50 x 10^17)

[M+2] = 0.160 /3.70 x 10^13

[M+2] = 4.32 x 10^-15

Concentration of M+2 at equilibrium is 4.32 x 10^-15 M

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