What is the equilibrium concentration of CN ion after mixing the following solutions? (Hint: Mixing the solutions dilutes the reactants.)(a) mixing 50.0 mL of 0.100 M HCN with 50.0 mL of 0.100 M NH3?Kc = 0.714.0✔NH3(aq) + HCN(aq) ⇒ NH4+ (aq) + CN ¯(aq)M(b) mixing 50.0 mL of 0.100 M HCN with 50.0 mL of 0.100 M K₂S?S² (aq) + HCN(aq) = HS¯(aq) + CN −(aq)Kc = 3.1 x 10³XM4.0 5.0e2(c) mixing 50.0 mL of 0.100 M HCN with 50.0 mL of 0.100 M KF?F (aq) + HCN(aq) = HF(aq) + CN ¯(aq)Kc = 5.6 x 10-7M4.0 3.7e-5

Answered: Image /qna-images/answer/fa18c124-93b3-4314-a97f-4afdf1770db8.jpg

(a) mixing 50.0 mL of 0.100 M HCN with 50.0 mL of 0.100 M NH3? NH3(aq) + HCN(aq) ⇒ NH4+ (aq) + CN ¯(aq) Kc = 0.71 M 4.0 (b) mixing 50.0 mL of 0.100 M HCN with 50.0 mL of 0.100 M K₂S? S² (aq) + HCN(aq) = HS¯(aq) + CN −(aq) Kc = 3.1 x 10³ 5.0e2 XM 4.0

(a) mixing 50.0 mL of 0.100 M HCN with 50.0 mL of 0.100 M NH3? NH3(aq) + HCN(aq) ⇒ NH4+ (aq) + CN ¯(aq) Kc = 0.71 M 4.0 (b) mixing 50.0 mL of 0.100 M HCN with 50.0 mL of 0.100 M K₂S? S² (aq) + HCN(aq) = HS¯(aq) + CN −(aq) Kc = 3.1 x 10³ 5.0e2 XM 4.0

Chemistry for Engineering Students

4th Edition

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Lawrence S. Brown, Tom Holme

Chapter12: Chemical Equilibrium

Section: Chapter Questions

Problem 12.25PAE

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Question

100%

What is the equilibrium concentration of CN ion after mixing the following solutions? (Hint: Mixing the solutions dilutes the reactants.)
(a) mixing 50.0 mL of 0.100 M HCN with 50.0 mL of 0.100 M NH3?
Kc = 0.71
4.0✔
NH3(aq) + HCN(aq) ⇒ NH4+ (aq) + CN ¯(aq)
M
(b) mixing 50.0 mL of 0.100 M HCN with 50.0 mL of 0.100 M K₂S?
S² (aq) + HCN(aq) = HS¯(aq) + CN −(aq)
Kc = 3.1 x 10³
XM
4.0 5.0e2
(c) mixing 50.0 mL of 0.100 M HCN with 50.0 mL of 0.100 M KF?
F (aq) + HCN(aq) = HF(aq) + CN ¯(aq)
Kc = 5.6 x 10-7
M
4.0 3.7e-5

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