Q10P Chemical equilibrium and analysi... [FREE SOLUTION]

Chapter 19: Q10P (page 486)

Chemical equilibrium and analysis of a mixture. (Warning! This is a long problem.) A remote optical sensor for ${CO}_{2}$ in the ocean was designed to operate without the need for calibration.33

The sensor compartment is separated from seawater by a silicone membrane through which ${CO}_{2}$ , but not dissolved ions, can diffuse. Inside the sensor, ${CO}_{2}$ equilibrates with ${HCO}_{3}$ and ${CO}_{3}^{2}$ . For each
measurement, the sensor is flushed with fresh solution containingbromothymol blue indicator. All indicator is in the formnear neutral pH, so we can
write two mass balances:
$[{HIn}^{鈭�}] + [\ln^{2 鈭�}] = F_{In} = 50 .0 渭惭补苍诲 [{Na}^{+}] = F_{狈伪} = 50 .0 渭惭 + 42 .0 渭惭 = 92 .0 渭惭$
has an absorbance maximum at 434 nm andhas a maximum at 620 nm. The sensor measures the absorbance ratio $R_{A} = A_{620} / A_{434}$ reproducibly without need for calibration. From this ratio, we can findin the seawater as outlined here:
(a).From Beer鈥檚 law for the mixture, write equations forin terms of the absorbance at 620 and 434 nmThen show that
$\frac{[\ln^{2 鈭�}]}{[Hln 鈭�]} = \frac{R_{A} 蔚_{434}^{{HHn}^{鈭�}} 鈭� 蔚_{6, 20}^{Hln 鈭�}}{蔚_{620}^{\ln^{2 鈭�}} 鈭� R_{A} 蔚_{434}^{\ln^{2 鈭�}}} = R_{\ln}$ (A)
(b) From the mass balance (1) and the acid dissociation constant
, show that
$[{Hln}^{鈭�}] = \frac{F_{1 n}}{R_{\ln} + 1}$ (B)
$[\ln^{2 鈭�}] = \frac{K_{\ln} F_{\ln}}{[H^{+}] (R_{\ln} + 1)}$ (C)
(c) Show that $[H^{+}] = K_{\ln} / R_{\ln}$ (D)
(d) From the carbonic acid dissociation equilibria, show that
$\begin{aligned} [{HCO}_{3}^{鈭�}] = \frac{K_{1} [CO (aq)] E}{[H^{+}]} \\ [{CO}_{3}^{2 鈭�}] = \frac{K_{1} K_{2} [CO (aq)] F}{{[H^{+}]}^{2}} \end{aligned}$
(e) Write the charge balance for the solution in the sensor compartment. Substitute in expressions B, C, E, and F forHln, ${In}^{2 -}, [{HCO}_{3}], and [{CO}_{3}^{2 鈭�}]$
(f) Suppose that the various constants have the following values:
$\begin{array}{ll} 蔚_{4344}^{{HHn}^{鈭�}} = 8 .00 脳 10^{3} M^{鈭� 1} {cm}^{鈭� 1} 鈥呪赌呪赌呪赌� K_{1} = 3 .0 脳 10^{鈭� 7} \\ 蔚_{6620}^{Hn 鈭�} = 0 鈥呪赌呪赌呪赌� K_{2} = 3 .3 脳 10^{鈭� 11} \\ 蔚_{434}^{\ln^{2}} = 1 .90 脳 10^{3} M^{鈭� 1} {cm}^{鈭� 1} 鈥呪赌呪赌呪赌� K_{\ln} = 2 .0 脳 10^{鈭� 7} \\ 蔚_{620}^{\ln^{2 鈭�}} = 1 .70 脳 10^{4} M^{鈭� 1} {cm}^{鈭� 1} 鈥呪赌呪赌呪赌� K_{w} = 6 .7 脳 10^{鈭� 15} \end{array}$
From the measured absorbance ratio=2.84, findin the seawater.
(g) Approximately what is the ionic strength inside the sensor compartment? Were we justified in neglecting activity coefficients in working this problem?

Short Answer

Expert verified

$\begin{aligned} (a) . Proved \frac{[{In}^{2 鈭�}]}{[{HIn}^{鈭�}]} = \frac{R_{A} 蔚_{434}^{{Hln}^{鈭�}} 鈭� 蔚_{6, 20}^{{Hln}^{鈭�}}}{蔚_{620}^{\ln_{2}^{2}} 鈭� R_{A} 蔚_{434}^{\ln^{2}}} = R_{\ln} \\ (b) . Proved [{HIn}^{鈭�}] = \frac{F_{in}}{R_{In} + 1} and [{In}^{2 鈭�}] = \frac{K_{\ln} F_{\ln}}{[H^{+}] (R_{\ln} + 1)} \\ (c) . Proved [H^{+}] = K_{\ln} / R_{\ln} \\ (d) . Proved [{HCO}_{3}^{鈭�}] = \frac{K_{1} [{CO}_{2} (aq)]}{[H^{+}]} and [{CO}_{3}^{2 鈭�}] = \frac{K_{1} K_{2} [{CO}_{2} (aq)]}{{[H^{+}]}^{2}} \end{aligned}$

(e). The charge balance is

$F_{Na} + [H^{+}] = \frac{K_{w}}{[H^{+}]} + \frac{F_{\ln}}{R_{\ln + 1}} + 2 \frac{K_{\ln} F_{lo}}{[H^{+}] + (R_{\ln + 1)}} + \frac{K_{1} [{CO}_{2 (xax)}]}{[H^{+}]} + 2 \frac{K_{1} K_{2} [{CO}_{2 (man}]}{{[H^{+}]}_{2}}$

(f) $[{CO}_{2 ((aq)}] = 3.04 脳 10^{鈭� 6} M$

(g). The ionic strength is $125 渭惭$ .

Step by step solution

Show that [In2−][HIn−]=RAε434Hnn−ε6,20Hn−2ε620ln2−RAε434ln2=RIn :

Consider,

A as absorbance,

e as molar absorptivity,

b as pathlength.

From the given question we can get,

$\begin{matrix} A_{620} = 蔚_{620}^{{HIn}^{鈭�}} b [{HIn}^{鈭�}] + 蔚 \ln_{620}^{2 鈭�} 鈦� b [\ln^{2 鈭�}] \\ A_{434} = e_{434}^{HIn} b [{HIn}^{鈭�}] + 蔚_{434}^{\ln 鈦� {nn}^{2}} b [\ln^{2 鈭�}] \end{matrix}$

By solving above equation we get,

$\begin{aligned} [{HInn}^{鈭�}] = \frac{1}{D} (A_{620} 蔚_{434}^{\ln^{2 鈭�}} b 鈭� A_{434} {In}_{620}^{\ln^{2 鈭�}} 鈦� b) \\ [\ln^{2 鈭�}] = \frac{1}{D} (A_{434} {HIn}_{620}^{鈭�} 鈦� b 鈭� A_{620434} \underset{Hn}{鈭� 1} b) \end{aligned}$

Divide ${In}^{2 -}$ by Hln ,

Divide $A_{434}$ to the numerator and denominator,

Hence proved that $\frac{[\ln^{2 鈭�}]}{[{HIn}^{鈭�}]} = \frac{R_{A} s_{434}^{Hn} 鈭� 蔚_{6, 20}^{HIn}}{蔚_{620}^{\ln^{2 鈭�}} 鈭� R_{A} s_{434}^{\ln^{2 鈭�}}} = R_{\ln}$

Show that [Hln−]=F1nRln+1 and [ln2−]=KlnFln[H+](Rln+1) :

$[{HIn}^{鈭�}] = \frac{F_{\ln}}{R_{mn + 1}}$

To prove $K_{In}$ ,

For indicator, the mass balance can be given as,

$[{HIn}^{鈭�}] + [\ln^{2 鈭�}] = F_{In}$

Divide Hln both side,

$\begin{matrix} \frac{{Hln}^{鈭�} 鈭�}{[{Hln}^{鈭�}]} + \frac{[\ln^{2}]}{[{Hln}^{鈭�}]} = \frac{F_{\ln}}{[{HIn}^{鈭�}]} \\ 1 + R_{\ln} = \frac{F_{\ln}}{[HI n^{鈭�}]} \end{matrix}$

Hence proved $[{HIn}^{鈭�}] = \frac{F_{\ln}}{R_{\ln + 1}}$

The acid dissociation constant of the indicator is,

$K_{\ln} = \frac{[\ln^{2}] [H^{+}]}{[{HInn}^{鈭�}]}$

Substitute $F_{In} / (R_{In} + 1) for [{HIn}^{鈭�}]$ ,

$\begin{matrix} K_{\ln} = \frac{[{In}^{2 鈭�}] [H^{+}] (R_{\ln} + 1)}{F_{In}} \\ [{In}^{2 鈭�}] = \frac{K_{\ln} F_{\ln}}{[H^{+}] (R_{\ln} +)} \end{matrix}$

Show that H+=Kln/Rln :

$R_{\ln} = \frac{[\ln^{2}]}{[H \ln^{鈭�}]}$

Hence, $K_{\ln} = \frac{[\ln^{2}] [H^{+}]}{[{HIn}^{鈭�}]}$

Therefore it has been proved, $[H^{+}] = K_{\ln} / R_{in}$ .

Show that [HCO3]=K1[CO(aq) ][H+]and [CO32−]=K1K2CO(aq) ][H+]2:

$K_{1} = \frac{[{HCO}_{3}^{鈭�}] [H^{+}]}{[{CO}_{2 (aq)}]}$ the acid dissociation reaction of Carbonic acid.

Hence proved $[{HCO}_{3}] = \frac{K_{1} [{CO}_{2 (aq)}]}{[H^{+}]}$

$K_{2} = \frac{[{CO}_{3}^{2 鈭�}] [H^{+}]}{[{HOO}_{3}^{鈭�}]}$ the acid dissociation reaction of Bicarbonate,

$[{CO}_{3}^{2 鈭�}] = \frac{K_{2} [{HCO}_{3}^{鈭�}]}{[H^{+}]}$

Substitute $[{HCO}_{3}]$ we get,

$[{CO}_{3}^{2 鈭�}] = \frac{K_{1} K_{2} [{CO}_{2 (aq)}]}{{[H^{+}]}^{2}}$

Find charge balance:

$\begin{aligned} Substitute the expression {Hin}^{鈭�}, {In}^{2 鈭�}, [{HCO}_{3}], and [{CO}_{3}^{2 鈭�}] . \\ [{Na}^{+}] + [H^{+}] = [{OH}^{鈭�}] + [{HIn}^{鈭�}] + 2 [{In}^{2 鈭�}] + [{HCO}_{3}^{鈭�}] + 2 [{CO}_{3}^{2 鈭�}] \\ The solution is \\ F_{Na} + [H^{+}] = \frac{K_{w}}{[H^{+}]} + \frac{F_{\ln}}{R_{\ln + 1}} + 2 \frac{K_{\ln} F_{in}}{[H^{+}] + (R_{\ln} + 1)} + \frac{K_{1} [{CO}_{2 (xq)}]}{[H^{+}]} + 2 \frac{K_{1} K_{2} [{CO}_{2 (mpq}]}{{[H^{+}]}_{2}} \end{aligned}$

find CO2⁡(aq) :

$\begin{aligned} [H^{+}] = K_{in} / R_{in} \\ For R_{in} \\ R_{in} = \frac{R_{A} {蔚_{434}}^{Hn} 鈭� 蔚_{620} H^{HIn}}{蔚_{620}^{1^{2 鈭�}} 鈭� R_{A} 蔚_{434} {In}^{2}} \\ = \frac{(2.84) (8.00 脳 10^{3}) 鈭� (0)}{(1.70 脳 10^{4}) 鈭� (2.84) (1.90 脳 10^{3})} \\ = 1.958 \\ [H^{+}] = (2.0 脳 10^{鈭� 7}) / 1.958 \\ [H^{+}] = 1.02 脳 10^{鈭� 7} M \end{aligned}$

To find $[{CO}_{2} (aq)]$ , The value of $[H^{+}]$ is substituted into mass balance

$F_{Na} + [H^{+}] = \frac{K_{w}}{[H^{+}]} + \frac{F_{in}}{R_{\ln}} + 2 \frac{K_{\ln} F_{in}}{[H^{+}] (R_{\ln} + 1)} + \frac{K_{1} [{CO}_{2 la})]}{[H^{+}]} + 2 \frac{K_{1} K_{2} [{CO}_{2 (aq)}]}{{[H^{+}]}^{2}}$

$\begin{aligned} 92 .0 脳 10^{鈭� 6} + 1 .02 脳 10^{鈭� 7} = \frac{(6 .7 脳 10^{鈭� 15})}{(1 .02 脳 10^{鈭� 7})} + \frac{(50 .0 脳 10^{6})}{1 .958 + 1} + 2 \frac{(2 .0 脳 10^{7}) (50 .0 脳 10^{6})}{(1 .02 脳 10^{鈭� 7}) (1 .958 + 1)} \\ + \frac{(3 .0 脳 10^{7}) [{CO}_{2 (aj}]}{(1 .02 脳 10^{鈭� 7})} + 2 \frac{(3 .0 脳 10^{7}) (3 .3 脳 10^{鈭� 11}) [{CO}_{2 x}]}{{(1 .02 脳 10^{鈭� 7})}^{1}} \\ 9 .21 脳 10^{鈭� 5} = 6 .56 脳 10^{鈭� 8} + 1 .69 脳 10^{鈭� 5} + 6 .62 脳 10^{鈭� 5} + 2 .94 [{CO}_{2 (aq)}] + 0 .0019 [{CO}_{2 (aq)}] \\ The Solution is [{CO}_{2 ((aq))}] = 3 .04 脳 10^{鈭� 6} M \end{aligned}$

Find ionic strength:

The ${Na}^{+}, {HIn}^{鈭�}, {In}^{2 鈭�}, {HCO}_{3}^{鈭�}, {CO}_{3}^{2 鈭�} and {OH}^{鈭�}$ are present in the solution. If the net cation charge is $92.1 渭惭$ ,the net charge on anion should be $92.1 渭惭$ and the ionic strength must approximately $~ 92 渭惭鈮� 10^{- 4} M$ activity coefficients are close to 1.00 and hence an ionic strength of $10^{- 4}$ Mis low .

$\begin{aligned} [{OH}^{鈭�}] = \frac{K_{w}}{[H^{+}]} = 0.07 渭 M \\ [{HIn}^{鈭�}] = \frac{F_{\ln}}{R_{\ln} + I} \\ = 16.9 渭 m \\ [\ln^{2 鈭�}] = \frac{K_{\ln} F_{\ln}}{[H^{+}] (R_{\ln} + 1)} \\ = 33 鈰� 1 渭 m \\ [{HCO}_{3}^{鈭�}] = \frac{K_{1} [{CO}_{2 (aq)}]}{[H^{+}]} \\ = 2.94 [{CO}_{2 (aq)}] \\ = 8.9 渭 M \\ [{CO}_{3}^{2 鈭�}] = \frac{K_{1} K_{2} [{CO}_{2 (mq)}]}{{[H^{+}]}^{2}} \\ = 0.0019 [{CO}_{2 (aq)}] \\ = 0.003 渭 M \end{aligned}$

$\begin{matrix} 鈭� c_{i} Z_{i}^{2} = \\ \{[{Na}^{+}] 鈰� 1^{2} + [H^{+}] 鈰� 1^{2} + [{OH}^{鈭�}] 鈰� 1^{2} + [{In}^{2 鈭�}] 鈰� 2^{2} + [{HCO}_{3}^{鈭�}] 鈰� 1^{2} + [{CO}_{3}^{2 鈭�}] 鈰� 2^{2} + [{CO}_{3}^{2 鈭�}] 鈰� 2^{2} \\ } \end{matrix}$

The solution for ionic strength is 125 $渭惭$ .

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Short Answer

Step by step solution

Show that [In2−][HIn−]=RAε434Hnn−ε6,20Hn−2ε620ln2−RAε434ln2=RIn :

Show that [Hln−]=F1nRln+1 and [ln2−]=KlnFln[H+](Rln+1) :

Show that H+=Kln/Rln :

Show that [HCO3]=K1[CO(aq) ][H+]and [CO32−]=K1K2CO(aq) ][H+]2:

Find charge balance:

find CO2⁡(aq) :

Find ionic strength:

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