Doświadczenie 1. Wyznaczenie pojemności buforowej roztworu buforowego (HA/A⁻; BOH, B⁺)

Stężenia roztworów używanych w doświadczeniu

C⁰_CH3COOH= 0,20 mol/dm³ C⁰_CH3COONa= 0,20 mol/dm³

C⁰_HCl = 0,25 mol/dm³ C⁰_NaOH = 0,25 mol/dm³

pK_a=4,75

pK_b=4,76

pK_w=14,00

Objętość roztworów użytych do sporządzenia roztworu buforowego:

V_CH3COOH = 20cm³ V_CH3COONa = 20cm³

Pomiar pojemności buforowej w stosunku do HCl:

Lp	VHCL	nHCL	Vr-r	Ca	Cs	pHobl	pHpom
	[cm^3]	[mmol]	[cm^3]	[mol/dm^3]	[mol/dm^3]
0	0	0	20	0, 1	0, 1	4, 75	4,68
1	1	0,25	21	0, 1071	0, 0833	4, 64	4,26
2	2	0,5	22	0, 1136	0, 0682	4, 53	4,11
3	3	0,75	23	0, 1196	0, 0543	4, 41	3,96
4	4	1	24	0, 1250	0, 0417	4, 27	3,81
5	5	1,25	25	0, 1300	0, 0300	4, 11	3,65
6	6	1,5	26	0, 1346	0, 0192	3, 9	3,45
7	7	1,75	27	0, 1389	0, 0093	3, 57	3,25
8	8	2	28	0, 1429	0	2, 80	2,98
9	9	2,25	29	0, 1466	0	2, 79	2,50
10	10	2,5	30	0, 1500	0	2, 78	1,89

Przykładowe obliczenia:

n_HCl= C_HCl*V_HCl,  np.:

0) n_HCl = 0, 25 mol/dm3  *  0 dm3 =  0

1) n_HCl = 0, 25 mol/dm3  *  0, 001 dm3 =  0,25

n_a^′= n_a⁰+ n_HCl

$$\mathbf{C}_{\mathbf{a}}\mathbf{= \ }\frac{\mathbf{C}_{\mathbf{a}}^{\mathbf{0}}\mathbf{*\ }\mathbf{V}_{\mathbf{a}}\mathbf{+}\mathbf{n}_{\mathbf{\text{HCl}}}}{V_{r - r}}\mathbf{,\ \ np.:}$$

$${0)\ C}_{a} = \frac{0,2\ \text{mol}/\text{dm}^{3}*0,01\ \text{dm}^{3}}{0,02\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1}$$

$${1)\ C}_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 0,25*10^{- 3}\ \text{mol}}{0,021\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1071}$$

n_s^′= n_s⁰−n_HCl

$$\mathbf{C}_{\mathbf{s}}\mathbf{= \ }\frac{\mathbf{C}_{\mathbf{s}}^{\mathbf{0}}\mathbf{*\ }\mathbf{V}_{\mathbf{s}}\mathbf{-}\mathbf{n}_{\mathbf{\text{HCl}}}}{V_{r - r}}\mathbf{,}\mathbf{\ \ np.:}$$

$${0)\ C}_{s} = \frac{0,2\ \text{mol}/\text{dm}^{3}*0,01\ \text{dm}^{3}}{0,02\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{1}$$

$$1)\ C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 0,25*10^{- 3}\ \text{mol}}{0,021\ \text{dm}^{3}} = \mathbf{0}\mathbf{,}\mathbf{0833}$$

$${8)\ C}_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 2,00*10^{- 3}\ \text{mol}}{0,028\ \text{dm}^{3}} = \mathbf{0}$$

PH_obl=  − log([H⁺])

$\mathbf{\text{gdy\ }}\mathbf{n}_{\mathbf{\text{HCl}}}\mathbf{<}\mathbf{n}_{\mathbf{s}}^{\mathbf{0}}\mathbf{\ ,\ \ \ \ \ \ }\left\lbrack \mathbf{H}^{\mathbf{+}} \right\rbrack\mathbf{= \ }\frac{\mathbf{Ka*\ }\mathbf{C}_{\mathbf{s}}^{\mathbf{'}}}{\mathbf{C}_{\mathbf{a}}^{\mathbf{'}}}\mathbf{\text{\ \ }}\mathbf{\text{\ \ }}\frac{\mathbf{Ka*}\mathbf{n}_{\mathbf{s}}^{\mathbf{'}}}{\mathbf{n}_{\mathbf{a}}^{\mathbf{'}}}$

$$\mathbf{\text{gdy\ }}\mathbf{n}_{\mathbf{s}}^{\mathbf{0}}\mathbf{= 0}\mathbf{,\ \ \ \ \ \ \ }\left\lbrack \mathbf{H}^{\mathbf{+}} \right\rbrack\mathbf{= \ }\sqrt{\mathbf{Ka*}\mathbf{C}_{\mathbf{a}}^{\mathbf{\ '}}\mathbf{\text{\ \ }}}\mathbf{\ ,\ \ \ \ \ \ np.:}$$

$$0)\ - \log\left( 10^{- 4,75}*\frac{0,1\ mol/\text{dm}^{3}}{0,1\ mol/\text{dm}^{3}} \right) = \mathbf{4,75}$$

$$1)\ - \log\left( 10^{- 4,75}*\frac{0,1071\ mol/\text{dm}^{3}}{0,0833\ mol/\text{dm}^{3}} \right) = \mathbf{4,64}$$

$$8)\ \ - \log\left( \sqrt{10^{- 4,75}*0,1429\ mol/\text{dm}^{3}} \right) = \mathbf{2,8}0$$

Pomiar pojemności buforowej w stosunku do NaOH:

Lp	VNAOH	nNAOH	Vr-r	Ca (Cb)	Cs	pHobl	pHpom
	[cm^3]	[mmol]	[cm^3]	[mol/dm^3]	[mol/dm^3]
0	0	0	20,0	0, 1	0, 1	4,75	4,19
1	1	0,25	21,0	0, 0833	0, 1071	4,86	4,27
2	2	0,5	22,0	0, 0682	0, 1136	4,97	4,46
3	3	0,75	23,0	0, 0543	0, 1196	5,09	4,54
4	4	1	24,0	0, 0417	0, 1250	5,23	4,70
5	5	1,25	25,0	0, 0300	0, 1300	5,39	4,90
6	6	1,5	26,0	0, 0192	0, 1346	5,60	5,17
7	7	1,75	27,0	0, 0093	0, 1389	5,93	5,78
8	8	2	28,0	0	0, 1429	8,95	12,09

Przykładowe obliczenia:

n_NaOH= C_NaOH*V_NaOH,  np.

0)  n_NaOH = 0, 25 mol/dm3  *  0 dm3 =  0

1)  n_NaOH = 0, 25 mol/dm3  *  0, 001 dm3 =  0, 25

n_a^′= n_a⁰− n_NaOH

$$\mathbf{C}_{\mathbf{a}}\mathbf{= \ }\frac{\mathbf{C}_{\mathbf{a}}^{\mathbf{0}}\mathbf{*\ }\mathbf{V}_{\mathbf{a}}\mathbf{-}\mathbf{n}_{\mathbf{\text{NaOH}}}}{V_{r - r}}\mathbf{,\ \ np.:}$$

$${0)\ \ C}_{a} = \frac{0,2\ mol/\text{dm}^{3}*0,01\ \text{dm}^{3}}{0,02\ \text{dm}^{3}} = \mathbf{0,1}$$

$$1)\ \ \ C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 0,25*10^{- 3}\text{\ mol}}{0,021\ \text{dm}^{3}} = \mathbf{0,0833}$$

$$8)\ \ \ C_{a} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) - 2,00*10^{- 3}\text{\ mol}}{0,028\ \text{dm}^{3}} = \mathbf{0}$$

n_s^′= n_s⁰+n_NaOH

$$\mathbf{C}_{\mathbf{s}}\mathbf{= \ }\frac{\mathbf{C}_{\mathbf{s}}^{\mathbf{0}}\mathbf{*\ }\mathbf{V}_{\mathbf{s}}\mathbf{+}\mathbf{n}_{\mathbf{\text{NaOH}}}}{V_{r - r}}\mathbf{,}\mathbf{\ \ np.:}$$

$${0)\ \ C}_{s} = \frac{0,2\ mol/\text{dm}^{3}*0,01\ \text{dm}^{3}}{0,02\ \text{dm}^{3}} = \mathbf{0,1}$$

$$1)\ \ C_{s} = \frac{\left( 0,2\frac{\text{mol}}{\text{dm}^{3}}*0,01\ \text{dm}^{3} \right) + 0,25*10^{- 3}\text{\ mol}}{0,021\ \text{dm}^{3}} = \mathbf{0,1071}$$

PH_obl=  − log([H⁺])

$\mathbf{\text{gdy\ }}\mathbf{n}_{\mathbf{\text{NaOH}}}\mathbf{<}\mathbf{n}_{\mathbf{a}}^{\mathbf{0}}\mathbf{\ ,\ \ \ \ \ \ }\left\lbrack \mathbf{H}^{\mathbf{+}} \right\rbrack\mathbf{= \ }\frac{\mathbf{Ka*\ }\mathbf{C}_{\mathbf{s}}^{\mathbf{'}}}{\mathbf{C}_{\mathbf{a}}^{\mathbf{'}}}\mathbf{\text{\ \ }}\mathbf{\text{\ \ }}\frac{\mathbf{Ka*}\mathbf{n}_{\mathbf{s}}^{\mathbf{'}}}{\mathbf{n}_{\mathbf{a}}^{\mathbf{'}}}$

$$\mathbf{\text{gdy\ }}\mathbf{n}_{\mathbf{\text{NaOH}}}\mathbf{=}\mathbf{n}_{\mathbf{a}}^{\mathbf{0}}\mathbf{,\ \ \ \ \ \ \ }\left\lbrack \mathbf{\text{OH}}^{\mathbf{-}} \right\rbrack\mathbf{= \ }\sqrt{\mathbf{Kh*}\mathbf{C}_{\mathbf{s}}^{\mathbf{\ '}}\mathbf{\text{\ \ }}}\mathbf{\ ,\ \ }\mathbf{Kh =}\frac{\mathbf{\text{Kw}}}{\mathbf{\text{Ka}}}\mathbf{\ \ \ \ np.:}$$

$$\operatorname{0)-log}\left( 10^{- 4,75}*\frac{0,1\ mol/\text{dm}^{3}}{0,1\ mol/\text{dm}^{3}} \right) = \mathbf{4,75}$$

$$1)\ \ - \log\left( 10^{- 4,75}*\frac{0,0833\ mol/\text{dm}^{3}}{0,1071\ mol/\text{dm}^{3}} \right) = \mathbf{4,86}$$

$$8)\ \ pOH = - \log\left( \sqrt{\frac{10^{- 14}}{10^{- 4,75}}*0,1429\ mol/\text{dm}^{3}} \right) = 5,05$$

pH = 14 − 5, 05 = 8, 95

Zależność pH roztworu buforowego od objętości dodanego

HCl / NaOH:

Obliczenie pojemności buforowej

a) w stosunku do HCl:

$$\beta = \frac{- n_{\text{HCl}}}{pH*V} =$$

b) w stosunku do NaOH:

$$\beta = \frac{n_{\text{NaOH}}}{pH*V} =$$