Explain why we can simplify the calculations for K when the value of K is very small (less4.1than -10*).4.2HCN will react very slightly with water with a K = 6.2 x 10-10.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 ofH;O*?. Then use the concentration of H3O* to calculate the pH of the solution with the equationpH = -log[H3O*] (you can do this!!!). Put your final answers in the box. If you need more spaceto show your work, you can use the backside of thispage.[H;O'] at equilibriumpH = -log(H,O]pH =% of HCN thatreacted

Answered: Image /qna-images/answer/48e93853-de3c-4778-bcff-897910ca3378.jpg

4.2 HCN will react very slightly with water with a K = 6.2 x 10-10. %3D 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,0*?. Then use the concentration of H,O* to calculate the pH of the solution with the equation pH = -log[H3O*] (you can do this!!!). Put your final answers in the box. If you need more space to show your work, you can use the backside of this page. [H,O'] at equilibrium M pH = -log(H,O'] pH = % of HCN that reacted

4.2 HCN will react very slightly with water with a K = 6.2 x 10-10. %3D 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,0?. Then use the concentration of H,O to calculate the pH of the solution with the equation pH = -log[H3O*] (you can do this!!!). Put your final answers in the box. If you need more space to show your work, you can use the backside of this page. [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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Consider the system 4NH3(g)+3O2(g)2N2(g)+6H2O(l)H=1530.4kJ (a) How will the concentration of ammonia at equilibrium be affected by (1) removing O2(g)? (2) adding N2(g)? (3) adding water? (4) expanding the container? (5) increasing the temperature? (b) Which of the above factors will increase the value of K? Which will decrease it?
Consider the system 4NH3(g)+3O2(g)2N2(g)+6H2O(l)H=1530.4kJ (a) How will the amount of ammonia at equilibrium be affected by 1. removing O2(g)? 2. adding N2(g)? 3. adding water? 4. expanding the container at constant pressure? 5. increasing the temperature? (b) Which of the above factors will increase the value of K? Which will decrease it?
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