Tampilkan postingan dengan label chemistry class 11. Tampilkan semua postingan
Tampilkan postingan dengan label chemistry class 11. Tampilkan semua postingan
Ionisation constant of CH3COOH is 1,7×10⁻⁵and concentration of H+ ions is 3,4×10-4. Then find out initial concentration of CH3COOH molecule ?
Ionisation constant of CH3COOH is 1,7×10⁻⁵and concentration of H+ ions is 3,4×10-4. Then find out initial concentration of CH3COOH molecule ?
SOLUTION
Known:
CH3COOH = 1,7x10-5
H+ = 3,4×10-4
Asked: concentration of CH3COOH molecule ?
CH3COOH —> CH3COO⁻ + H+
x 3,4x10-4 3x10-4
K = [CH3COO-] [H+]/[CH3COOH]
1,7x10-5 = [3,4x10-4] [3,4x10-4]/x
x = 6,8x10-3
so, the concentration of CH3COOH molecule is 6,8x10-3
A solution consists of 0.2M NH4OH and 0.2M NH4Cl. If Kb of NH4OH is 1.8 × 10^5, the [OH-] of the resulting solution is ?
A solution consists of 0.2M NH4OH and 0.2M NH4Cl. If Kb of NH4OH is 1,8x10-5, the [OH-] of the resulting solution is ?
A. 3,6x10-5
B. 3,2x10-5
D. 1,8x10-5
C. 0,9x10-5
SOLUTION
NH4OH and NH4Cl is a buffer solution:
POH = PKb + log [salt]/[base]
Concentration of NH4OH = 0,2 M
Concentration of NH4Cl = 0,2 M
PKb of NH4OH = 1,8x10-5
POH = - log 1,8x10-5 + log [0,2]/[0,2]
POH = - log 1,8x10-5
-log [OH-] = - log 1,8x10-5
[OH-] = 1,8x10-5
well, the concentration of [OH-] is 1,8x10-5
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