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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
200 ml of 1 M H2SO4, 300 ml of 3 M HCl and 100 ml of 2 M HCl are mixed and made up to 1 litre. The proton concentration in the resulting solution is:
200 ml of 1 M H2SO4, 300 ml of 3 M HCl and 100 ml of 2 M HCl are mixed and made up to 1 litre. The proton concentration in the resulting solution is ?
A. 0,25 M
B. 1,25 M
C. 2,5 M
D. 1,5 M
E. 0,75 M
SOLUTION
H2SO4 contains two H+ ions
So 200 ml of H2SO4 contains 0,2L×1mol/L×2 = 0,4 mol H+
similarly 300 ml of HCL of 3M contains 0,3 L × 3 mol/L × 1 = 0,9 mol H+
similarly 100 ml of HCL of 2M contains 0,1L × 2 mol/L × 1 = 0,2 mol H+
Asked: The proton concentration in the resulting solution is ?
Total number of moles = 0,4 + 0,9 + 0,2 = 1,5 mol
Total volume = 200 ml + 300 mL + 100mL = 600mL => 0,6 L
And Concentration of
[H+] = 1,5/0,6 mol/L = 2,5 mol/L
So, The proton concentration in the resulting solution is 2,5 mol/L.
How many moles of NH3CI must be added to 1.5 L of 0.2 M solution of NH3 to form a buffer whose PH is 9.00 (Kb = 1.8 x 10-3) ?
How many moles of NH3CI must be added to 1.5 L of 0.2 M solution of NH3 to form a buffer whose PH is 9.00 (Kb = 1.8 x 10-3) ?
SOLUTION
Known:
V = 1,5 L
PH = 9.00
NH3 = 0,2 M
Kb = 1.8 x 10-3
Asked: How many moles of NH3CI ?
Answer:
pOH = pKb + log (Csalt/Cbase)
14 - pH = -log Kb + log (Csalt/Cbase)
14 - 9 = -log (1.8x10-3) + log (Csalt/0.2)
52.745 + log (Csalt/0.2)
log (Csalt/0.2) = 2.255
Csalt/0.2 = 5.555x10-3
Csalt = 0.2 x 5.555x10-3
Csalt = 1.111 x 10-3 M
NH3 = V . Csalt
NH3 = 1,5 . 1.111x10-3 M
NH3 = 1.667×10-3 M
So, How many moles of NH3CI must be added to 1,5 L of 0.2 M is 1.667×10-3 M.
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
3. An aqueous solution contains 0.01M RNH₂ ( Kb = 2x10-6 ) & 10-4 M NaOH. The concentration of OH- is nearly ?
3. An aqueous solution contains 0.01M RNH₂ ( Kb = 2x10-6 ) & 10-4 M NaOH. The concentration of OH- is nearly ?
A. 2.414×10-4
B. 2×10-4
C. 1.414×10-5
D. 10-4
E. 1.414×10-4
SOLUTION
[OH−] ⟹ RNH2 is weak base, it reacts with H2O
pKb = −log10 (2×10-6)
= 6 − log2
pH = 14 − 21 (pKb − logC)
pH = 14 − 21 (6-log2 − log10−2)
pH = 14 − 3 + 2 log2 + 21(−2)
pH = 10 + 2 log2
pH = 10.150515
NaOH ⟹ [OH−]= 10-4 M
pOH = 14 − pH
pOH = 0.0001414 ⟹ 1.414×10-4 M
So, the correct concentration of OH- is 1.414×10-4 M (E)
5. A solution contains 0.2M NH4OH and 0.2M NH4CI. If 1.0 mL of 0.001 M HCl is added to it. What will be the [OH-] of the resulting solution [Kb = 2x10-5] ?
5. A solution contains 0.2M NH4OH and 0.2M NH4CI. If 1.0 mL of 0.001 M HCl is added to it. What will be the [OH-] of the resulting solution [Kb = 2x10-5] ?
A. 2x10-5
B. 2x10-3
C. 5x10-10
D. None of these
SOLUTION
Concentration of NH4OH = 0.2M
Concentration of NH4Cl = 0.2M
Concentration of HC1 = 0.001 M
[Kb = 2x10-5]
Applying Henderson Equation
POH = PKa + log [salt/base]
POH = 5 - log 2 + log(0.2/0.2)
POH = 5 - log2
[OH-] =2x10-5
So, the [OH-] of the resulting solution is 2x10-5 (the option (A) is the correct answer).
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