Basic Buffer Action
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Q. NH4OH is titrated with HCl solution. Find the value of log([NH+4][NH4OH]) when the volume of HCl added is half of that of equivalence point.
- 0
- 1
- 0.5
- 0.75
Q. Calculate the weight (in mg) of HCl that should be added to 100 ml of 0.1 M BOH to make a buffer solution of pOH=6.
Kb(BOH)=1×10−8 (assume there is no change in volume on addition of HCl)
Kb(BOH)=1×10−8 (assume there is no change in volume on addition of HCl)
- 4.6
- 3.6
- 5.6
- 6.6
Q. Calculate the weight in mg of HCl added to 100 ml of 0.1 N BOH to have its pH = 6.6 and Kb=6.25×10−8; Antilog (−7.4)=3.98×10−8. (Assume there is no change in volume on addition of HCl)
- 300
- 425
- 100
- 223
Q. Which will make basic buffer ?
- 50 mL of 0.1M NaOH+25 mL of 0.1M CH3COOH
- 100 mL of 0.1 M CH3COOH+100 mL of 0.1 M NaOH
- 100 mL of 0.1 M HCl+200 mL of 0.1M NH4OH
- 100 mL of 0.1 M HCl+100 mL of 0.1 M NaOH
Q. Calculate the weight (in mg) of HCl that should be added to 100 ml of 0.1 M BOH to make a buffer solution of pOH=6.
Kb(BOH)=1×10−8 (assume there is no change in volume on addition of HCl)
Kb(BOH)=1×10−8 (assume there is no change in volume on addition of HCl)
- 3.6
- 4.6
- 5.6
- 6.6
Q. Which will make basic buffer ?
- 50 mL of 0.1M NaOH+25 mL of 0.1M CH3COOH
- 100 mL of 0.1 M CH3COOH+100 mL of 0.1 M NaOH
- 100 mL of 0.1 M HCl+200 mL of 0.1M NH4OH
- 100 mL of 0.1 M HCl+100 mL of 0.1 M NaOH
Q. Which will make a basic buffer?
- 50 mL of 0.1 M NaOH + 25 mL of 0.1 M CH3COOH
- 100 mL of 0.1 M CH3COOH + 100 mL of 0.1 M NaOH
- 100 mL of 0.1 M HCl + 200 mL of 0.1 M NH4OH
- 100 mL of 0.1 M HCl + 100 mL of 0.1 M NaOH
Q. Conductometric titration curve of an equimolar mixture of HCl and HCN with NaOH (aq) can be given as
Q. CH3NH2 (0.1 mole Kb = 5 ×10−4) is added to 0.08 mole of HCl and the solution is diluted to one litre, resulting hydrogen ion concentration is
- 1.6×10−11
- 8×10−11
- 5×10−5
- 8×10−2
Q. NH4OH is titrated with HCl solution. Find the value of log([NH+4][NH4OH]) when the volume of HCl added is half of that of equivalence point.
- \N
- 0.5
- 0.75
- 1
Q. Calculate the weight in mg of HCl added to 100 ml of 0.1 N BOH to have its pH = 6.6 and Kb=6.25×10−8; Antilog (−7.4)=3.98×10−8. (Assume there is no change in volume on addition of HCl)
- 300
- 100
- 425
- 223
Q. Calculate the ratio of CH3NH2 to CH3NH+3Cl− required to create a buffer with pH = 10.14
Kb of CH3NH2=4.4×10−4
take log 2.27=0.356, log 0.313=−0.504
Kb of CH3NH2=4.4×10−4
take log 2.27=0.356, log 0.313=−0.504
- 0.5
- 1.05
- 0.31
- None of the above