作业帮Chemistry problems!Help!1.Calculate the pH when we add 0.001 mole of HAc to 0.02 mole of NaAc and bring the volume up to 100ml with pure water.HAc ↔ H+ + Ac− pKa = 4.762.Consider the following two systems:(a) 0.1 litre of pure water at pH
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Chemistry problems!Help!1.Calculate the pH when we add 0.001 mole of HAc to 0.02 mole of NaAc and bring the volume up to 100ml with pure water.HAc ↔ H+ + Ac− pKa = 4.762.Consider the following two systems:(a) 0.1 litre of pure water at pH
Chemistry problems!Help!
1.Calculate the pH when we add 0.001 mole of HAc to 0.02 mole of NaAc and bring the volume up to 100ml with pure water.
HAc ↔ H+ + Ac− pKa = 4.76
2.Consider the following two systems:
(a) 0.1 litre of pure water at pH 7.0
(b) 0.1 litre of 0.1M phosphate buffer at pH 7.2
H2PO4 ↔ HPO42− + H+ pKa = 7.2
If we add 10ml of 0.01 M HCl to each,what will be the pH in each solution.
p/s:Solve the questions,not translate it for me!Thanks!
I need the answer in English,please..Help!
Chemistry problems!Help!1.Calculate the pH when we add 0.001 mole of HAc to 0.02 mole of NaAc and bring the volume up to 100ml with pure water.HAc ↔ H+ + Ac− pKa = 4.762.Consider the following two systems:(a) 0.1 litre of pure water at pH
1.suppose [H+]=x mol/L
before ionization
[HAc]=0.01mol/L [Ac-]=0.2mol/L
HAc ↔ H+ + Ac−
before 0.01 0 0.2
after 0.01-x x 0.2+x
thus -lg[(0.2+x)*x/(0.01-x)]=4.76
note that x is quite small
therefore 0.2+x=0.2 0.01-x=0.01
-lg20x=4.76
pH=-lgx=6.06
2.HCl=H+ +Cl-
(a) [H+]=0.01*10/(10+100)=9.09*10^-4mol/L
pH=-lg[H+]=3.04
(b) suppose ionized[H+]=xmol/L
H2PO4 ↔ HPO42− + H+ pKa = 7.2
before 0.1/1.1 0.1/1.1 0.01/11
after 0.1/1.1-x 0.1/1.1+x 0.01/11+x
thus -lg[(0.01/11+x)*(0.1/1.1+x)/(0.1/1.1-x)]=7.2
note that x is quite small
thus 0.1/1.1+-x=0.1/1.1
pH=-lg(0.01/11+x)=7.2
我很想给你说答案。但是你太庄逼。
有点晕...我觉得你先发到英语区翻译一下再发这里来比较好...