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№1 слайд![BUFFER SOLUTIONS](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img0.jpg)
Содержание слайда: BUFFER SOLUTIONS
№2 слайд![Buffer solutions solution](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img1.jpg)
Содержание слайда: Buffer solutions
solution which can resist the addition of a strong acid or a strong base or water. Its’ pH changes very slightly.
+ 1 drop of base [H+] in 1000 000 times
+ 1 drop of acid [H+] in 5000 times
(from 10-7 tо 5 х10-4)
In buffer solution from 1.00х10-7
to 1.01х10-7
№3 слайд![Classification](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img2.jpg)
Содержание слайда: Classification
№4 слайд![](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img3.jpg)
№5 слайд![](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img4.jpg)
№6 слайд![HOW TO PREPARE BUFFER .](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img5.jpg)
Содержание слайда: HOW TO PREPARE BUFFER
1. Mixing the components:
-for acidic buffer
pH = pKa + lg Ns·Vs/Na·Va
-for basic buffer
pH = 14 – pKв – lg Ns·Vs/Nb·Vb
№7 слайд![. Partial neutralization For](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img6.jpg)
Содержание слайда: 2. Partial neutralization
For acidic buffer
nacid = nbase = nsalt
СН3СООН + NaOH = CH3COONa + H2O
(exsess)
pH = pKa + lg Nb·Vb / (Na·Va –Nb·Vb)
For basic buffer
NH4OH + HCL = NH4Cl + H2O
(exsess)
pH = 14 – pKв – lg Na·Va / (Nb·Vb - Na· Va)
№8 слайд![Buffer capacity Ba nacid р Н.](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img7.jpg)
Содержание слайда: Buffer capacity
Ba = nacid / |∆р Н|. Vbuf.sol
Вb = nbase/ |∆р Н|. Vbuf.sol
n – mole equivalents of a strong acid or a strong base
Vbuf.sol - volume of a buffer solution
∆рН – pH change as a result of acid or base addition
№9 слайд![Buffer capacity depends on pH](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img8.jpg)
Содержание слайда: Buffer capacity depends on :
pH = pKa + lg nsalt/nacid
pH = 14 - pKb - lg nsalt/nbase
№10 слайд![](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img9.jpg)
№11 слайд![Buffer capacity nsalt gt](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img10.jpg)
Содержание слайда: Buffer capacity
nsalt > nacid Вa > Вb
nsalt < nacid Вa < Вb
nsalt = nacid Вa = Вb =Вmax
pH = pKa + lg nsalt/nacid
№12 слайд![Choose the buffer with](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img11.jpg)
Содержание слайда: Choose the buffer with maximum capacity and рН = 7.36 :
Choose the buffer with maximum capacity and рН = 7.36 :
1) acetic рК = 4.75;
2) phosphate рК = 7.21;
3) hydrocarbonate рК = 6.37.
№13 слайд![Buffer systems of a body](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img12.jpg)
Содержание слайда: Buffer systems of a body
1.Mineral
Hydrocarbonate Н2СО3
НСО3–
Phosphate Н2РО4–
НРO42–
2. Protein and aminoacidic.
№14 слайд![](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img13.jpg)
№15 слайд![](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img14.jpg)
№16 слайд![. A buffer consists of ,](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img15.jpg)
Содержание слайда: 1. A buffer consists of 0,5 moles of equivalent NH3 and 0,5 moles of equivalent NH4Cl. Which buffer component must be added to change pH to 9? Kb(NH3)=1,8*10-5
2. What is the pH of buffer made of
60 ml of 0,10M NH3 with 40 ml of 0,10M NH4Cl. Kb=1,8*10-5.
3. What volume of 0,6M CH3COONa must be added to 600 ml of 0,2M CH3COOH to produce a buffer with pH=4,75? Ka(CH3COOH)=1,75*10-5.
№17 слайд![. What volume of , M NaOH](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img16.jpg)
Содержание слайда: 4. What volume of 0,01M NaOH should be added to 100 ml of 0,5M CH3COOH solution to produce a buffer with pH 4,75? pKa(CH3COOH)=4,75
5. A buffer was prepared of 500 ml NaН2РО4 and 500 ml Na2НРO4 . After addition of 1 ml 0.1N HCl the change of buffer pH = 0.03. Calculate buffer capacity Ba.
6. Choose a buffer with Вa > Вb:
a). 100 ml 0.2M NaHCO3 + 100ml 0.4M H2CO3
b). 100 ml 0.4M NaHCO3 + 100ml 0.2M H2CO3
c). 100 ml 0.2M NaHCO3 + 100ml 0.2M H2CO3
№18 слайд![](/documents_6/befb7b33d8a2c00368c8f5c7c9f38568/img17.jpg)