1) The substance diethylamine is a weak nitrogenous base like ammonia. Write a n
ID: 995923 • Letter: 1
Question
1) The substance diethylamine is a weak nitrogenous base like ammonia.
Write a net ionic equation to show how diethylamine, (C2H5)2NH, behaves as a base in water. state is its s/l/g/aq.
2) The substance trimethylamine is a weak nitrogenous base like ammonia.
Complete the following equation that shows how trimethylamine reacts when dissolved in water.
?
3)
Which of the following aqueous solutions are buffer solutions ?
.
0.33 M NH4NO3 + 0.40 M NH3
.
0.13 M HCN + 0.11 M KCN
.
0.25 M HBr + 0.20 M KBr
.
0.34 M CH3COOH + 0.23 M CH3COOK
.
0.16 M KOH + 0.24 M KBr
4)
Which of the following aqueous solutions are buffer solutions ?
.
0.29 M NH4Br + 0.35 M NH3
.
0.18 M NaOH + 0.26 M NaCl
.
0.27 M HNO3 + 0.19 M KNO3
.
0.40 M NaBr + 0.28 M NaNO3
.
0.10 M HF + 0.16 M KF
5)
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Explanation / Answer
1) (C2H5)2NH + H2O -----------> (C2H5)2NH2+ + OH-
2) (CH3)3N + H2O ------------> (CH3)3NH+ + OH-
3) Buffer solution = weak acid and its salt with strong base
Buffer solution = weak base and its salt with strong acid
0.33 M NH4NO3 + 0.40 M NH3, 0.13 M HCN + 0.11 M KCN, 0.34 M CH3COOH + 0.23 M CH3COOK
are buffer solutions .
Because, NH3 is a weak base and HCN,CH3COOH are weak acids.
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0.25 M HBr + 0.20 M KBr , 0.16 M KOH + 0.24 M KBr are not buffer solutions because HBr is a strong acid and KOH is a strong base.
4)
0.29 M NH4Br + 0.35 M NH3 , 0.10 M HF + 0.16 M KF are buffer solutions because NH3 is a weak base and HF is a weak acid.
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0.18 M NaOH + 0.26 M NaCl , 0.27 M HNO3 + 0.19 M KNO3, 0.40 M NaBr + 0.28 M NaNO3
are not buffer solutions because NaOH is a strong base and HNO3 is a strong acid.
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