Question 14.E.F: H2S from cigarette smoke was collected by bubbling smoke thr...
H_{2}S from cigarette smoke was collected by bubbling smoke through aqueous NaOH and measured with a sulfide ion-selective electrode. Standard additions of volume V_{S} containing Na_{2}S at concentration c_{S} = 1.78 mM were then made to V_{0} = 25.0 mL of unknown and the electrode response, E, was measured.
From a separate calibration curve, it was found that β = 0.985 in Equation 14-14. Using T = 298.15 K and n = −2 (the charge of S^{2−}), prepare a standard addition graph with Equation 14-15 and find the concentration of sulfide in the unknown.
E = k + β(\frac{RT \ln 10}{nF})\log[X] (14-14)
\underbrace{(V_{0} + V_{S})10^{E/S}}_{y} = \underbrace{10^{k/S}V_{0}c_{X}}_{b} + \underbrace{10^{k/S}c_{S}}_{m}\underset{\begin{matrix} \uparrow \\ x \end{matrix} }{V_{S}} (14-15)
| V_{S} (mL) | E (v) | V_{S} (mL) | E (v) |
| 0 | 0.046 5 | 3.00 | 0.030 0 |
| 1.00 | 0.040 7 | 4.00 | 0.026 5 |
| 2.00 | 0.034 4 |
The function to plot on the y-axis is (V_{0} + V_{S})10^{E/S} , where S = (βRT/nF) \ln 10. β is is 0.985. Putting in R = 8.314 5 J/mol · K), F = 96 485 C/mol, T = 298.15 K, and n = −2 gives S = −0.029 136 J/C = −0.029 136 V. (Recall that joule/coulomb = volt.)
The data are plotted in Figure 14-30, which has a slope of m = 0.744 84
and an intercept of b = 0.439 19, giving an x-intercept of −b/m = −0.58_{9 65} mL. The concentration of original unknown is
c_{X} = \frac{(x-intercept)c_{S}}{V_{0}} = −\frac{(−0.58_{9 65} mL)(1.78 mM)}{25.0 mL} = 4.2 × 10^{−5} M
(We decided that the last significant digit in the x-intercept was the 0.01 decimal place because the original data were only measured to the 0.01 decimal place.)
| V_{S} (mL) | E (v) | y |
| 0 | 0.046 5 | 0.633 8 |
| 1.00 | 0.040 7 | 1.042 5 |
| 2.00 | 0.034 4 | 1.781 1 |
| 3.00 | 0.030 0 | 2.615 2 |
| 4.00 | 0.026 5 | 3.571 7 |
