chemiabufory sprawozdanie

Podstawy chemii nieorganicznej

Temat ćwiczenia: Roztwory buforowe

Doświadczenie 1.

Wyznaczanie pojemności buforowej roztworu buforowego (HA/A-;BOH/B+)

Stężenia roztworów używanych w doświadczeniu

C0CH3COOH= 0,20 mol/dm3 C0CH3COONa= 0,20 mol/dm3

C0NH3*H2O= 0,20 mol/dm3 C0NH4Cl =0,20 mol/dm3

C0HCl = 0,25 mol/dm3 C0NaOH = 0,25 mol/dm3

pKa=4,75

pKb=4,76

pKw=14,00

Pomiar pojemności buforowej w stosunku do HCl:

Lp VHCL nHCL Vr-r Ca Cs pHobl pHpom
[cm3] [mmol] [cm3] [mol/dm3] [mol/dm3]
0 0 0 20
    0, 1
    0, 1
4, 75
 4,761
1 1 0,25 21
0, 1071
0, 0833
4, 64
 4,672
2 2 0,5 22
0, 1136
0, 0682
4, 53
 4,588
3 3 0,75 23
0, 1196
0, 0543
4, 41
 4,482
4 4 1 24
0, 1250
0, 0417
4, 27
 4,436
5 5 1,25 25
0, 1300
0, 0300
4, 11
 4,370
6 6 1,5 26
0, 1346
0, 0192
3, 9
 4,233
7 7 1,75 27
0, 1389
0, 0093
3, 57
 4,111
8 8 2 28
0, 1429
0
2, 80
 3,985
9 9 2,25 29
0, 1466
 0
2, 79
 3,821
10 10 2,5 30
0, 1500
 0
2, 78
 3,588

Wzory oraz przykładowe obliczenia

  1. nHCl= CHCl*VHCl

Lp [mmol]
0 nHCl=0,25 mol/dm3 * 0 dm3= 0
1 nHCl=0,25 mol/dm3 * 0,001 dm3= 0,25
2 nHCl=0,25 mol/dm3 * 0,002 dm3= 0,5
3 nHCl=0,25 mol/dm3 * 0,003 dm3= 0,75
4 nHCl=0,25 mol/dm3 * 0,004 dm3= 1
5 nHCl=0,25 mol/dm3 * 0,005 dm3= 1,25
6 nHCl=0,25 mol/dm3 * 0,006 dm3= 1,5
7 nHCl=0,25 mol/dm3 * 0,007 dm3= 1,75
8 nHCl=0,25 mol/dm3 * 0,008 dm3= 2
9 nHCl=0,25 mol/dm3 * 0,009 dm3= 2,25
10 nHCl=0,25 mol/dm3 * 0,01 dm3= 2,5

  1. Ca

Lp [mol/dm3]
0
$$C_{a} = \frac{0,2\ \text{mol}/\text{dm}^{3}*0,01\ \text{dm}^{3}}{0,02\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1}$$
1
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 0,25*10^{- 3}\ \text{mol}}{0,021\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1071}$$
2
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 0,50*10^{- 3}\ \text{mol}}{0,022\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1136}$$
3
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 0,75*10^{- 3}\ \text{mol}}{0,023\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1196}$$
4
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 1,00*10^{- 3}\ \text{mol}}{0,024\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1250}$$
5
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 1,25*10^{- 3}\ \text{mol}}{0,025\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1300}$$
6
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 1,5*10^{- 3}\ \text{mol}}{0,026\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1346}$$
7
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 1,75*10^{- 3}\ \text{mol}}{0,027\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1389}$$
8
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 2,00*10^{- 3}\ \text{mol}}{0,028\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1429}$$
9
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 2,25*10^{- 3}\ \text{mol}}{0,029\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1466}$$
10
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 2,50*10^{- 3}\ \text{mol}}{0,03\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1500}$$

  1. CS

Lp [mol/dm3]
0
$$C_{s} = \frac{0,2\ \text{mol}/\text{dm}^{3}*0,01\ \text{dm}^{3}}{0,02\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1}$$
1
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 0,25*10^{- 3}\ \text{mol}}{0,021\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{0833}$$
2
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 0,50*10^{- 3}\ \text{mol}}{0,022\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{0682}$$
3
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 0,75*10^{- 3}\ \text{mol}}{0,023\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{0543}$$
4
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 1,00*10^{- 3}\ \text{mol}}{0,024\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{0417}$$
5
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 1,25*10^{- 3}\ \text{mol}}{0,025\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{0300}$$
6
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 1,5*10^{- 3}\ \text{mol}}{0,026\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{0192}$$
7
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 1,75*10^{- 3}\ \text{mol}}{0,027\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{0093}$$
8
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 2,00*10^{- 3}\ \text{mol}}{0,028\ \text{dm}^{3}} = \mathbf{0}$$
9
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 2,25*10^{- 3}\ \text{mol}}{0,029\ \text{dm}^{3}}$$
10
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 2,50*10^{- 3}\ \text{mol}}{0,03\ \text{dm}^{3}}$$

  1. pHobl

Lp
0
$$- \log\left( 10^{- 4,75}*\frac{0,1\ mol/\text{dm}^{3}}{0,1\ mol/\text{dm}^{3}} \right) = \mathbf{4,75}$$
1
$$- \log\left( 10^{- 4,75}*\frac{0,1071\ mol/\text{dm}^{3}}{0,0833\ mol/\text{dm}^{3}} \right) = \mathbf{4,64}$$
2
$$- \log\left( 10^{- 4,75}*\frac{0,1136\ mol/\text{dm}^{3}}{0,0682\ mol/\text{dm}^{3}} \right) = \mathbf{4,53}$$
3
$$- \log\left( 10^{- 4,75}*\frac{0,1196\ mol/\text{dm}^{3}}{0,0543\ mol/\text{dm}^{3}} \right) = \mathbf{4,41}$$
4
$$- \log\left( 10^{- 4,75}*\frac{0,1250\ mol/\text{dm}^{3}}{0,0417\ mol/\text{dm}^{3}} \right) = \mathbf{4,27}$$
5
$$- \log\left( 10^{- 4,75}*\frac{0,1300\ mol/\text{dm}^{3}}{0,0300\ mol/\text{dm}^{3}} \right) = \mathbf{4,11}$$
6
$$- \log\left( 10^{- 4,75}*\frac{0,1346\ mol/\text{dm}^{3}}{0,0192\ mol/\text{dm}^{3}} \right) = \mathbf{3,9}$$
7
$$- \log\left( 10^{- 4,75}*\frac{0,1389\ mol/\text{dm}^{3}}{0,0093\ mol/\text{dm}^{3}} \right) = \mathbf{3,57}$$
8
$$- \log\left( \sqrt{10^{- 4,75}*0,1429\ mol/\text{dm}^{3}} \right) = \mathbf{2,8}0$$
9
$$- \log\left( \sqrt{10^{- 4,75}*0,1466\ mol/\text{dm}^{3}} \right) = \mathbf{2,79}$$
10
$$- \log\left( \sqrt{10^{- 4,75}*0,1500\ mol/\text{dm}^{3}} \right) = \mathbf{2,78}$$

Pomiar pojemności buforowej w stosunku do NaOH:

Lp VNAOH nNAOH Vr-r Ca (Cb) Cs pHobl pHpom
[cm3] [mmol] [cm3] [mol/dm3] [mol/dm3]
0 0 0 20,0
    0, 1
    0, 1
 4,75 4,723
1 1 0,25 21,0
0, 0833
0, 1071
 4,86 4,839
2 2 0,5 22,0
0, 0682
0, 1136
 4,97 4,956
3 3 0,75 23,0
0, 0543
0, 1196
 5,09 5,085
4 4 1 24,0
0, 0417
0, 1250
 5,23 5,231
5 5 1,25 25,0
0, 0300
0, 1300
 5,39 5,410
6 6 1,5 26,0
0, 0192
0, 1346
 5,60 5,654
7 7 1,75 27,0
0, 0093
0, 1389
 5,93 6,185
8 8 2 28,0
0
0, 1429
 8,95 11,293

Wzory oraz przykładowe obliczenia

  1. nNaOH= CNaOH*VNaOH

Lp [mmol]
0 nNaOH=0,25 mol/dm3 * 0 dm3= 0
1 nNaOH=0,25 mol/dm3 * 0,001 dm3= 0,25
2 nNaOH=0,25 mol/dm3 * 0,002 dm3= 0,5
3 nNaOH=0,25 mol/dm3 * 0,003 dm3= 0,75
4 nNaOH=0,25 mol/dm3 * 0,004 dm3= 1
5 nNaOH=0,25 mol/dm3 * 0,005 dm3= 1,25
6 nNaOH=0,25 mol/dm3 * 0,006 dm3= 1,5
7 nNaOH=0,25 mol/dm3 * 0,007 dm3= 1,75
8 nNaOH=0,25 mol/dm3 * 0,008 dm3= 2

  1. Ca

Lp [mol/dm3]
0
$$\text{\ \ \ \ \ C}_{a} = \frac{0,2\ mol/\text{dm}^{3}*0,01\ \text{dm}^{3}}{0,02\ \text{dm}^{3}} = \mathbf{0,1}$$
1
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 0,25*10^{- 3}\text{\ mol}}{0,021\ \text{dm}^{3}} = \mathbf{0,0833}$$
2
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 0,50*10^{- 3}\text{\ mol}}{0,022\ \text{dm}^{3}} = \mathbf{0,0682}$$
3
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 0,75*10^{- 3}\text{\ mol}}{0,023\ \text{dm}^{3}} = \mathbf{0,0543}$$
4
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 1,00*10^{- 3}\text{\ mol}}{0,024\ \text{dm}^{3}} = \mathbf{0,0417}$$
5
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 1,25*10^{- 3}\text{\ mol}}{0,025\ \text{dm}^{3}} = \mathbf{0,0300}$$
6
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 1,5*10^{- 3}\text{\ mol}}{0,026\ \text{dm}^{3}} = \mathbf{0,0192}$$
7
$$C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 1,75*10^{- 3}\text{\ mol}}{0,027\ \text{dm}^{3}} = \mathbf{0,0093}$$
8
$$\text{\ \ \ }C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 2,00*10^{- 3}\text{\ mol}}{0,028\ \text{dm}^{3}} = \mathbf{0}$$

  1. CS

Lp [mol/dm3]
0
$$C_{s} = \frac{0,2\ mol/\text{dm}^{3}*0,01\ \text{dm}^{3}}{0,02\ \text{dm}^{3}} = \mathbf{0,1}$$
1
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 0,25*10^{- 3}\text{\ mol}}{0,021\ \text{dm}^{3}} = \mathbf{0,1071}$$
2
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 0,50*10^{- 3}\text{\ mol}}{0,022\ \text{dm}^{3}} = \mathbf{0,1136}$$
3
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 0,75*10^{- 3}\text{\ mol}}{0,023\ \text{dm}^{3}} = \mathbf{0,1196}$$
4
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 1,00*10^{- 3}\text{\ mol}}{0,024\ \text{dm}^{3}} = \mathbf{0,1250}$$
5
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 1,25*10^{- 3}\text{\ mol}}{0,025\ \text{dm}^{3}} = \mathbf{0,1300}$$
6
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 1,5*10^{- 3}\text{\ mol}}{0,026\ \text{dm}^{3}} = \mathbf{0,1346}$$
7
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 1,75*10^{- 3}\text{\ mol}}{0,027\ \text{dm}^{3}} = \mathbf{0,1389}$$
8
$$C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 2,00*10^{- 3}\text{\ mol}}{0,028\ \text{dm}^{3}} = \mathbf{0,1429}$$

  1. pHobl

Lp
0
$$- \log\left( 10^{- 4,75}*\frac{0,1\ mol/\text{dm}^{3}}{0,1\ mol/\text{dm}^{3}} \right) = \mathbf{4,75}$$
1
$$- \log\left( 10^{- 4,75}*\frac{0,0833\ mol/\text{dm}^{3}}{0,1071\ mol/\text{dm}^{3}} \right) = \mathbf{4,86}$$
2
$$- \log\left( 10^{- 4,75}*\frac{0,0682\ mol/\text{dm}^{3}}{0,1136\ mol/\text{dm}^{3}} \right) = \mathbf{4,97}$$
3
$$- \log\left( 10^{- 4,75}*\frac{0,0543\ mol/\text{dm}^{3}}{0,11963\ mol/\text{dm}^{3}} \right) = \mathbf{5,09}$$
4
$$- \log\left( 10^{- 4,75}*\frac{0,0417\ mol/\text{dm}^{3}}{0,1250\ mol/\text{dm}^{3}} \right) = \mathbf{5,23}$$
5
$$- \log\left( 10^{- 4,75}*\frac{0,0300\ mol/\text{dm}^{3}}{0,1300\ mol/\text{dm}^{3}} \right) = \mathbf{5,39}$$
6
$$- \log\left( 10^{- 4,75}*\frac{0,0192\ mol/\text{dm}^{3}}{0,1346\ mol/\text{dm}^{3}} \right) = \mathbf{5,60}$$
7
$$- \log\left( 10^{- 4,75}*\frac{0,0093\ mol/\text{dm}^{3}}{0,1389\ mol/\text{dm}^{3}} \right) = \mathbf{5,93}$$
8


$$p_{\text{OH}} = - \log\left( \sqrt{\frac{10^{- 14}}{10^{- 4,75}}*0,1429\ mol/\text{dm}^{3}} \right) = 5,05$$


pH = 14 − 5, 05 = 8, 95

Zależność pH roztworu buforowego od objętości dodanego

HCl / NaOH:

Obliczenie pojemności buforowej

a) w stosunku do HCl:

b) w stosunku do NaOH:


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