Explain why we can simplify the calculations for K when the value of K is very small (less than ~104). 4.1 4.2 HCN will react very slightly with water with a K = 6.2 x 1010. HCN(aq) + H,O() H,O* (aq) and CN¯ (aq) A chemist starts with a 0.50 M HCN solution. What is the equilibrium concentration of H;O*?. Then use the concentration of H3O* to calculate the pH of the solution with the equation pH = -log[H,O*] (you can do this!!!). Put your final answers in the box. If you need more space to show your work, Nou can use the backside of this page. %3D [H,O'] at equilibrium M pH = -log[H,O*] pH = % of HCN that reacted

Explain why we can simplify the calculations for K when the value of K is very small (less than ~104). 4.1 4.2 HCN will react very slightly with water with a K = 6.2 x 1010. HCN(aq) + H,O() H,O* (aq) and CN¯ (aq) A chemist starts with a 0.50 M HCN solution. What is the equilibrium concentration of H;O?. Then use the concentration of H3O to calculate the pH of the solution with the equation pH = -log[H,O] (you can do this!!!). Put your final answers in the box. If you need more space to show your work, Nou can use the backside of this page. %3D [H,O'] at equilibrium M pH = -log[H,O] pH = % of HCN that reacted

Principles of Modern Chemistry

8th Edition

ISBN:9781305079113

Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler

Chapter16: Solubility And Precipitation Equilibria

Section: Chapter Questions

Problem 70AP

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