Principle of Electroneutrality
- Page ID
- 199366
Q1. List the cations and anions that were determined this term
This answer will differ based on the type of sample available, time spent on the project, etc.In any case, the answer to this question will guide the student toward constructing the appropriate charge balance equation. For Lithia water, a typical list of ions is:
Cations: Na^{+}, K^{+}, Li^{+}, Ca^{2+}, Mg^{2+}, Fe^{2+}
Anions: HCO_{3}^{-}, Cl^{-}, BO_{2}^{-}, SO_{4}^{2-}, SiO_{3}^{2-}
Q2. Based on the principle of electroneutrality, what should be true about the concentration of cationic charge and anionic charge in Lithia water.
The number of moles of positive charge should equal the number of moles of negative charge.
Q3. Assuming that the inorganic constituents of Lithia water have been completely characterized, write down an appropriate charge balance equation for all dissolved ions in Lithia water.
Based on the list of ions in Q1 of this section, and assuming that silica exists in solution as the silicate ion, the appropriate charge balance equation for the inorganic ions in Lithia water would be:
\[\ce{[Na+] + [K+] + [Li+] + 2[Ca^2+] + 2[Mg^2+] + 2[Fe^2+]} = \ce{[HCO3- ] + [Cl- ] + [BO2- ] + [SO4^2- ] + [SiO3^2- ]}\nonumber\]
Because the pH of Lithia water is 6.4 and the concentration of the other ions is large, the contribution of H^{+} and OH^{-} to the charge balance reaction can be neglected. In other cases, such as solutions having low ionic strength, it may be necessary to include these species in the calculation of electroneutrality.
Q4. Use the error propagation equations described in Chapter 4 of Analytical Chemistry 2.0 to propagate the experimental error in the following analytical determinations
- A solution with a volume of 1.000 ± 0.012 mL was determined to have a mass of 1.008 ± 0.023 g. What is the density of the solution and its uncertainty?
\[D= \dfrac{M}{V}= \dfrac{1.008\: g}{1.000\: mL}=1.008\: g\: mL^{-1}\nonumber\]
To estimate the uncertainty of an experimental result that is calculated using solely multiplication and division:
\[\dfrac{u_D}{D}= \sqrt{\left(\dfrac{u_M}{M}\right)^2 + \left(\dfrac{u_V}{V}\right)^2 }\nonumber\]
\[\dfrac{u_D}{1.008\: g\: mL^{-1} }= \sqrt{\left(\dfrac{0.023\: g}{1.008\: g}\right)^2 + \left(\dfrac{0.012\: mL}{1.000\: mL}\right)^2 }\nonumber\]
\[u_D=0.026\: g\: mL^{-1}\nonumber\]
The density of the solution is 1.008 ± 0.026 g mL^{-1}.
- The volume delivered by a buret (∆V) is the difference between the initial volume (V_{i}) and the final volume (V_{f}) read on the scale of the buret. If V_{i} is (5.36 ± 0.02) mL and V_{f} is (15.68 ± 0.02) mL, calculate the volume delivered and its uncertainty.
\[\mathrm{∆V = V_f - V_i = 15.68\: mL - 5.36\: mL = 10.32\: mL}\nonumber\]
To estimate the uncertainty of an experimental result that is calculated using solely addition and subtraction:
\[u_{∆V}=\sqrt{u_{V_f}^2 + u_{V_i}^2 }\nonumber\]
\[u_{∆V}=\sqrt{(0.02\: mL)^2+(0.02\: mL)^2 }=0.028\: mL\nonumber\]
Rounded to the nearest hundredth of a milliliter, the delivered volume is (10.32 ± 0.03) mL.
- The density of a liquid was determined by measuring the mass of a 100-mL volumetric flask filled with the liquid, subtracting the mass of the empty 100-mL volumetric flask, and dividing the mass difference by the volume of the 100-mL volumetric flask. If the mass of the full flask is 248.3 ± 0.1 g, the mass of the empty flask is 45.5 ± 0.1 g, and the volume of the flask is 100.0 ± 0.08 mL, calculate the density of the liquid and its uncertainty.
\[D_{liq}=\dfrac{(M_{full}-M_{empty} )}{V_{flask}} = \dfrac{(248.3\: g-45.5\: g)}{100.0\: mL}=2.028\: g\: mL^{-1}\nonumber\]
To estimate the uncertainty of an experimental result that is calculated using two or more sets of mathematical operations (in this case, addition and/or subtraction followed by multiplication and/or division), one must propagate error through one set of operations (in this case, subtraction of two masses) followed by propagating the error through the remaining set of operations (in this case, dividing volume into mass):
\[(M_{full}-M_{empty} )=(248.3\: g-45.5\: g)= 202.8\: g\nonumber\]
\[u_{∆M}=\sqrt{u_{M_{full}}^2 + u_{M_{empty}}^2 }\nonumber\]
\[u_{∆V}=\sqrt{(0.1\: g)^2+(0.1\: g)^2} =0.141\: g \nonumber\]
(keep additional 1 or 2 digits to minimize rounding errors)
\[\dfrac{u_D}{D}= \sqrt{\left(\dfrac{u_M}{M}\right)^2 + \left(\dfrac{u_V}{V}\right)^2} \nonumber\]
\[\dfrac{u_D}{2.028\: g\: mL^{-1} }= \sqrt{\left(\dfrac{0.141\: g}{202.8\: g}\right)^2 + \left(\dfrac{0.08\: mL}{100.0\: mL}\right)^2 }\nonumber\]
\[u_D=0.002\: g\: mL^{-1}\nonumber\]
The density of the liquid is 2.028 ± 0.002 g mL^{-1}.
Q5. Based on your reading of Chapter 4 in Analytical Chemistry 2.0, what statistical approach would you use to determine whether equivalent amounts of cationic charge and anionic charge were determined in Lithia water?
A t-test for two experimental means would be the appropriate statistical test for determining whether statistically equivalent amounts of cationic charge and anionic charge were determined in Lithia water.
Q6. The Nutrition Facts label on a bottle of local mineral water contains the following information on select inorganic solutes (standard deviations provided by author, assume each result based on triplicate readings (n=3)):
Solute |
Concentration (mg L^{-1}) |
---|---|
Bicarbonate |
466 (± 24) |
Magnesium |
124 (± 5) |
Calcium |
2.67 (± 0.10) |
Sodium |
3.31 (± 0.15) |
Zinc |
4.45 (± 0.10) |
At the 95% confidence level, determine whether all of the major inorganic solutes have been reported on the Nutrition Facts label from this bottle of mineral water.
Step 1: Convert all solute concentrations and uncertainties into molarities
\[\left(\dfrac{466\: mg\: HCO_3^-}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: HCO_3^-}{61.02\: g\: HCO_3^-} \right)=7.637 \times 10^{-3}\: M\: HCO_3^-\nonumber\]
\[\left(\dfrac{24\: mg\: HCO_3^-}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: HCO_3^-}{61.02\: g\: HCO_3^- }\right)=0.393 \times 10^{-3}\: M\: HCO_3^-\nonumber\]
\[\left(\dfrac{124\: mg\: Mg^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Mg^{2+}}{24.305\: g\: Mg^{2+}} \right)=5.102 \times 10^{-3}\: M\: Mg^{2+}\nonumber\]
\[\left(\dfrac{5\: mg\: Mg^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Mg^{2+}}{24.305\: g\: Mg^{2+}} \right)=0.206 \times 10^{-3}\: M\: Mg^{2+}\nonumber\]
\[\left(\dfrac{2.67\: mg\: Ca^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Ca^{2+}}{40.08\: g\: Ca^{2+}} \right)=6.662 \times 10^{-5}\: M\: Ca^{2+}\nonumber\]
\[\left(\dfrac{0.10\: mg\: Ca^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Ca^{2+}}{40.08\: g\: Ca^{2+}}\right)=0.250 \times 10^{-5}\: M\: Ca^{2+}\nonumber\]
\[\left(\dfrac{3.31\: mg\: Na^+}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Na^+}{22.9898\: g\: Na^+ }\right)=1.440 \times 10^{-4}\: M\: Na^+\nonumber\]
\[\left(\dfrac{0.15\: mg\: Na^+}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Na^+}{22.9898\: g\: Na^+ }\right)=0.065 \times 10^{-4}\: M\: Na^+\nonumber\]
\[\left(\dfrac{4.45\: mg\: Zn^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Zn^{2+}}{65.37\: g\: Zn^{2+} }\right)=6.807 \times 10^{-5}\: M\: Zn^{2+}\nonumber\]
\[\left(\dfrac{0.10\: mg\: Zn^{2+}}{L}\right)\left(\dfrac{1\: g}{1000\: mg}\right)\left(\dfrac{1\: mol\: Zn^{2+}}{65.37\: g\: Zn^{2+}}\right)=0.153 \times 10^{-5}\: M\: Zn^{2+}\nonumber\]
Solute |
Concentration (mg L^{-1}) |
Concentration (M x 10^{3}) |
---|---|---|
Bicarbonate |
466 (± 24) |
(7.637 ± 0.393) |
Magnesium |
124 (± 5) |
(5.102 ± 0.206) |
Calcium |
2.67 (± 0.10) |
(0.066 ± 0.002) |
Sodium |
3.31 (± 0.15) |
(0.144 ± 0.007) |
Zinc |
4.45 (± 0.10) |
(0.068 ± 0.002) |
Step 2: Create a charge balance equation for the bottled spring water sample based on the nutrition label.
\[[HCO_3^- ]=2[Mg^{2+} ]+ 2[Ca^{2+} ]+ [Na^+ ]+ 2[Zn^{2+} ]\nonumber\]
Step 3: Use the charge balance equation and error propagation to determine the total cationic charge and total anionic charge including their associated uncertainties.
Anionic Charge: 7.637 mM
Anionic Charge Uncertainty: \(s_-=0.393\: mM\nonumber\)
(bicarbonate ion is the only anionic species in this charge balance equation)
Cationic Charge: 2(5.102 mM) + 2(0.066 mM) + (0.144 mM) + 2(0.068 mM) = 10.616 mM
Cationic Charge Uncertainty:
\[s_+= \sqrt{(2(0.206\: mM))^2+ (2(0.0025\: mM))^2+ (0.0065\: mM)^2+ (2(0.0015\: mM))^2 }\nonumber\]
\[s_+=0.412\: mM\nonumber\]
Step 4: Use the F-test to see if the variance for the anionic charge is significantly different than the variance of the cationic charge. If so, the t-test for two experimental means with equal variances can be used to compare the concentrations of the anionic charge and cationic charge. If not, the t-test for two experimental means with unequal variances must be used to compare the concentrations of the anionic charge and cationic charge.
\[F_{exp}= \dfrac{(0.412\: mM)^2}{(0.393\: mM)^2} =1.10\nonumber\]
At 95% confidence level: \(F_{crit} (0.05,2,3)=16.04\)
degrees of freedom: \(ν_{num}=2\)
(variance in numerator based on triplicate determinations of bicarbonate ion
\[ν_{denom}=3 \nonumber\]
(there are four cationic analytes)
Since \(F_{exp}<F_{crit}\),
the variances of the two groups are statistically equivalent @ 95% confidence level
Step 5: Use the t-test for two experimental means with equal variances to determine if the anionic and cationic solute concentrations are statistically equivalent at the 95% confidence level.
\[t_{exp}= \dfrac{|\bar{X}_+ - \bar{X}_- |}{s_{pooled} \sqrt{\dfrac{1}{n_+} +\dfrac{1}{n_-} }}\nonumber\]
where \(s_{pooled}= \sqrt{\dfrac{(n_+-1) s_+^2 + (n_--1) s_-^2}{(n_++ n_-- 2)}}\)
\[s_{pooled}= \sqrt{\dfrac{(4-1) (0.412\: mM)^2+(3-1) (0.393\: mM)^2}{(4+ 3 - 2) }}=0.405\: mM\nonumber\]
\[t_{exp}= \dfrac{|10.616\: mM-7.637\: mM|}{(0.405\: mM)\sqrt{\dfrac{1}{4}+\dfrac{1}{3}}}=9.63\nonumber\]
\(t_{critical} (0.05,5)=2.571\), and therefore, \(t_{expt}>t_{critical}\)
the difference in the ionic concentrations is statistically significant at the 95% c.l.