Calculate the pH of a buffer prepared by mixing 300 cc of 0,3 M NH3 and 500 cc of 0,5 M NH4CI. Kb for NH3 = 1,8 × 10^-5SOLUTIONKnown:NH3 = 0,3 MV NH3 = 300 ccNH4Cl = 0,5 MV NH4Cl = 500 ccAsked: Calculate the pH of a buffer ?Answer:Given, kb NH3 = 1.8x10^-5Number of moles of NH3 =...
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 ?SOLUTIONKnown:CH3COOH = 1,7x10-5H+ = 3,4×10-4Asked: concentration of CH3COOH molecule ?CH3COOH —> CH3COO⁻ + H+x 3,4x10-4 3x10-4K = [CH3COO-] [H+]/[CH3COOH]1,7x10-5 = [3,4x10-4] [3,4x10-4]/xx = 6,8x10-3so, the concentration of CH3COOH molecule...
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 MB. 1,25 MC. 2,5 M D. 1,5 M E. 0,75 MSOLUTIONH2SO4 contains two H+ ions So 200 ml of H2SO4 contains 0,2L×1mol/L×2 = 0,4...
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) ?SOLUTIONKnown:V = 1,5 LPH = 9.00NH3 = 0,2 MKb = 1.8 x 10-3Asked: How many moles of NH3CI ?Answer:pOH = pKb + log (Csalt/Cbase)14 - pH = -log Kb +...
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. 3,6x10-5B. 3,2x10-5D. 1,8x10-5C. 0,9x10-5SOLUTIONNH4OH and NH4Cl is a buffer solution:POH = PKb + log [salt]/[base]Concentration of NH4OH = 0,2 MConcentration of NH4Cl = 0,2 MPKb of NH4OH = 1,8x10-5 POH = - log 1,8x10-5 +...
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. 1.414×10-5D. 10-4 E. 1.414×10-4 SOLUTION[OH−] ⟹ RNH2 is weak base, it reacts with H2OpKb = −log10 (2×10-6) = 6 − log2pH = 14 − 21 (pKb − logC)pH = 14 − 21 (6-log2 − log10−2)pH = 14 − 3...
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