Question 8.E.A: Using activity coefficients correctly, find the pH of 1.0 × ...
Using activity coefficients correctly, find the pH of 1.0 × 10^{−2} M NaOH.
pH = −\log \mathcal{A}_{H^{+}} . But \mathcal{A}_{H^{+}} \mathcal{A}_{OH^{−}} = K_{w} ⇒ \mathcal{A}_{H^{+}} = K_{w}/\mathcal{A}_{OH^{−}}.
For 1.0 × 10^{−2} M NaOH, [OH^{−}] = 1.0 × 10^{−2} M and \gamma _{OH^{−}} = 0.900
(Table 7-1, with μ = 0.010 M).
| TABLE 7-1 Activity coefficients for aqueous solutions at 25°C | ||||||
|
Ion |
Ion
size (α , pm) |
Ionic strength (μ , M) |
||||
|
0.001 |
0.005 |
0.01 |
0.05 |
0.1 |
||
| Charge = ± 1 |
Activity coefficient (\gamma ) |
|||||
| H^{+} |
900 |
0.967 | 0.933 | 0.914 | 0.86 |
0.83 |
| (C_{6}H_{5})_{2}CHCO_{2}^{-} , (C_{3}H_{7})_{4}N^{+} |
800 |
0.966 | 0.931 | 0.912 | 0.85 |
0.82 |
| (O_{2}N)_{3}C_{6}H_{2}O^{-} , (C_{3}H_{7})_{3}NH^{+} , CH_{3}OC_{6}H_{4}CO_{2}^{-} , |
700 |
0.965 | 0.930 | 0.909 | 0.845 |
0.81 |
| Li^{+} , C_{6}H_{5}CO_{2}^{-} , HOC_{6}H_{4}CO_{2}^{-} , ClC_{6}H_{4}CO_{2}^{-} , C_{6}H_{5}CH_{2}CO_{2}^{-} , CH_{2}==CHCH_{2}CO_{2}^{-} , (CH_{3})_{2}CHCH_{2}CO_{2}^{-} , (CH_{3}CH_{2})_{4}N^{+} , (C_{3}H_{7})_{2}NH_{2}^{+} |
600 |
0.965 | 0.929 | 0.907 | 0.835 |
0.80 |
| Cl_{2}CHCO_{2}^{-} , Cl_{3}CCO_{2}^{-} , (CH_{3}CH_{2})_{3}NH^{+} , (C_{3}H_{7})_{2}NH_{3}^{+} |
500 |
0.964 | 0.928 | 0.904 | 0.83 |
0.79 |
| Na^{+} , CdCl^{+} , ClO_{2}^{-} , IO_{3}^{-} , HCO_{3}^{-} , H_{2}PO_{4}^{-} , HSO_{3}^{-} , H_{2}AsO_{4}^{-} , CO(NH_{3})_{4}(NO_{2})_{2}^{+} , CH_{3}CO_{2}^{−} , ClCH_{2}CO_{2}^{−} , (CH_{3})_{4}N^{+} , (CH_{3}CH_{2})_{2}NH_{2}^{+} , H_{2}NCH_{2}CO_{2}^{-} |
450 |
0.964 | 0.928 | 0.902 | 0.82 |
0.775 |
| ^{+}H_{3}NCH_{2}CO_{2}H, (CH_{3})_{3}NH^{+} , CH_{3}CH_{2}NH_{3}^{+} |
400 |
0.964 | 0.927 | 0.901 | 0.815 |
0.77 |
| OH^{-} , F^{−} , SCN^{−} , OCN^{−} , HS^{−} , ClO_{3}^{−} , ClO_{4}^{−} , BrO_{3}^{−} , IO_{4}^{−} , MnO_{4}^{−} , HCO_{2}^{−} , H_{2citrate}^{−} , CH_{3}NH_{3}^{+} , (CH_{3})_{2}NH_{2}^{+} |
350 |
0.964 | 0.926 | 0.900 | 0.81 |
0.76 |
| K^{+} , Cl^{−} , Br^{−} , I^{−} , CN^{−} , NO_{2}^{−} , NO_{3}^{−} |
300 |
0.964 | 0.925 | 0.899 | 0.805 |
0.755 |
| Rb^{+} , Cs^{+} , NH_{4}^{+} , Tl^{+} , Ag^{+} |
250 |
0.964 | 0.924 | 0.898 | 0.80 |
0.75 |
| Charge = ±2 |
Activity coefficient ( \gamma ) |
|||||
| Mg^{2+} , Be^{2+} |
800 |
0.872 | 0.755 | 0.69 | 0.52 |
0.45 |
| CH_{2}(CH_{2}CH_{2}CO_{2}^{−})_{2} , (CH_{2}CH_{2}CH_{2}CO_{2}^{−})_{2} |
700 |
0.872 | 0.755 | 0.685 | 0.50 |
0.425 |
| Ca^{2+} , Cu^{2+} , Zn^{2+} , Sn^{2+} , Mn^{2+} , Fe^{2+} , Ni^{2+} , Co^{2+} , C_{6}H_{4}(CO_{2}^{−})_{2} , H_{2}C(CH_{2}CO_{2}^{−})_{2} , (CH_{2}CH_{2}CO_{2}^{−})_{2} |
600 |
0.870 | 0.749 | 0.675 | 0.485 |
0.405 |
| Sr^{2+} , Ba^{2+} , Cd^{2+} , Hg^{2+} , S^{2−} , S_{2}O_{4}^{2−} , WO_{4}^{2−} , H_{2}C(CO_{2}^{−})_{2} , (CH_{2}CO_{2}^{−})_{2} , (CHOHCO_{2}^{−})_{2} |
500 |
0.868 | 0.744 | 0.67 | 0.465 |
0.38 |
| Pb^{2+} , CO_{3}^{2−} , SO_{3}^{2−} , MoO_{4}^{2−} , Co(NH_{3})_{5}Cl^{2+} , Fe(CN)_{5}NO^{2−} , C_{2}O_{4}^{2−} , Hcitrate^{2−} |
450 |
0.867 | 0.742 | 0.665 | 0.455 |
0.37 |
| Hg_{2}^{2+} , SO_{4}^{2−} , S_{2}O_{3}^{2−} , S_{2}O_{6}^{2−} , S_{2}O_{8}^{2−} , SeO_{4}^{2−} , CrO_{4}^{2−} , HPO_{4}^{2−} |
400 |
0.867 | 0.740 | 0.660 | 0.445 |
0.355 |
| Charge = ±3 |
Activity coefficient (\gamma ) |
|||||
| Al^{3+} , Fe^{3+} , Cr^{3+} , Sc^{3+} , Y^{3+} , In^{3+} , lanthanides^{a} |
900 |
0.738 | 0.54 | 0.445 | 0.245 |
0.18 |
| citrate^{3−} |
500 |
0.728 | 0.51 | 0.405 | 0.18 |
0.115 |
| PO_{4}^{3−} , Fe(CN)_{6}^{3−} , Cr(NH_{3})_{6}^{3+} , Co(NH_{3})_{5}H_{2}O^{3+} |
400 |
0.725 | 0.505 | 0.395 | 0.16 |
0.095 |
| Charge = ±4 |
Activity coefficient ( \gamma ) |
|||||
| Th^{4+} , Zr^{4+} , Ce^{4+} , Sn^{4+} |
1 100 |
0.588 | 0.35 | 0.255 | 0.10 |
0.065 |
| Fe(CN)_{6}^{4−} |
500 |
0.57 | 0.31 | 0.20 | 0.048 |
0.021 |
a. Lanthanides are elements 57–71 in the periodic table.
\mathcal{A}_{H^{+}} = \frac{K_{w}}{[OH^{−}]\gamma _{OH^{−}}} = \frac{1.0 × 10^{−14}}{(1.0 × 10^{−2})(0.900)}
=1.11 × 10^{−12} ⇒ pH = −\log \mathcal{A}_{H^{+}} = 11.95