hhah State University CHEM 1212 Principles of Chemistry 2 Spring 2018 6. Using t
ID: 1034520 • Letter: H
Question
hhah State University CHEM 1212 Principles of Chemistry 2 Spring 2018 6. Using the dissociation constant. Ka:7.8x10-18, calculate the equilibrium concentrations of Cd2+ and CN in a 0.250-M solution of Cd(CN) 7. Calculate [HgCl, ] in a solution prepared by adding 00200 mol of NaCI to 0.250 L of a 0.100-M HgCl2 solution. IN Torr,-1.2 × 10', 8. Calculate the Fe equilibrium concentration when 0.0888 mole of K Fe(CN] is added to a solution with 0.0.00010 MCN 9. Explain why a buffer can be prepared from a mixture of NH Cl and NaOH but not from NHs and NaOH 10. What is [H,O'] in a solution of 0.25 M CH CO:H and 0.030 M NaCH CO,? 1.8 x 10-5 Ka= H,O. (aq) + CH.co2-(aq) CHCO:H(aq) + H20( ) ?t 11. What will be the pH of a buffer solution prepared from 0.20 mol NHs, 0.40 mol NH ag) +H,0)NH, (a)OH (aq1.8x10 NHNOs, and just enough water to give 1.00 L of solution?Explanation / Answer
6. for the reaction,
Cd(CN)4^2- <==> Cd2+ + 4CN-
initial [Cd(CN)4^2-] = 0.250 M
Cd(CN)4^2- <==> Cd2+ + 4CN-
I 0.250 - -
C -x +x +4x
E 0.250-x x 4x
So,
Kd = [Cd2+][CN-]^4/[Cd(CN)4^2-]
7.8 10^-18 = (x)(4x)^4/(0.250-x)
256x^5 + 7.8 x 10^-18x - 1.95 x 10^-18 = 0
x = 8.73 x 10^-11 M
So the equilibrium concentration of,
[Cd2+] = 8.73 x 10^-11 M
[CN-] = 4 x 8.73 x 10^-11 = 3.50 x 10^-10 M
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7. For the reaction,
HgCl2 + 2NaCl <==> HgCl4^2-
Kf = 1.2 x 10^15
initial HgCl2 = 0.25 L x 0.1 M = 0.025 moles
initial NaCl = 0.02 moles
HgCl2 + 2NaCl <==> HgCl4^2-
I 0.025 0.020 -
C -x -2x +x
E 0.025-x 0.02-2x x
So,
1.2 x 10^15 = x/(0.025-x)(0.02-2x)^2
2.16 x 10^14x^2 - 3 x 10^12x + 1.2 x 10^10 - 4.8 x 10^15x^3 = 0
so,
[HgCl4^2-] = x = 0.007 moles
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9. a buffer is a mixture of weak acid and its conjugate base or a weak base and its conjugate acid. In NH4Cl (weak acid) and NaOH (strong base) we get NH4+/NH3 formation, which is a buffer.
Whereas, in NH3(weak base) and NaOH (strong base) we cannot form buffer with two bases. Their is no acid component in the buffer here.
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10. pH of buffer
using Hendersen-Haselbalck equation,
pH = pKa + log(NaCH3CO2-/CH3COOH)
feeding the given values,
pH = 4.74 + log(0.030/0.25) = 3.82
pH = -log[H3O+]
[H3O+] = 1.51 x 10^-4 M
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11. pH of buffer
using Hendersen-Haselbalck equation,
pH = pKa + log(NH3/NH4+)
feeding the given values,
pH = 10.25 + log(0.20/0.40) = 9.95
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