13. Below is the analysis of a natural water. Species Concentration (ppm) Na+ 15

Question

13. Below is the analysis of a natural water. Species Concentration (ppm) Na+ 150. K+ 10.0 Ca2+ 32.0 Mg2+ 27.0 SO4 2- 280 Sr2+ 0.400 Cl- 15.0 HCO3 - 150. SiO2(aq) 19.0 pH 7.40

(a) Convert the concentrations of the aqueous species from ppm (part per million) to molality (mol kg-1), except for pH, which should be expressed as the activity of H+. Show your work. You are welcome to use an Excel spreadsheet, and cut and paste your answer as a table. Would this be considered a "fresh" water or a "salt" water? Briefly explain why.

(b) Is this solution charge balanced (i.e., electrically neutral)? Calculate the total number of moles of positive charge and the total number of moles of negative charge (i.e., mi zi + i = mi zi i , where m is the molal concentration and is the charge on the ion, either positive or negative). Express the difference in charge as mol equivalents (eq) of charge / kg solution. Is there excess positive charge or negative charge?

(c) What is the percent charge imbalance in the solution, i.e. (eq mols charge imbalance) / (total eq mols pos+neg charge) * 100? Is this a large or small amount of charge imbalance? Briefly discuss reasons why a solution may not be exactly charged balanced.

Explanation / Answer

a) I would not put the table, instead I'll show you the work for the changes of concentration from ppm to molality.

Na = 150 mg/L or 150 mg/kg, (0.15 g/L) (This is posible because water has a density of 1, so kg and L are the same.)

MM = 23 g/mol

m = 0.15 / 23 = 6.52x10-3 mol/kg

K = 10 mg/kg or 0.010 g/kg; MM = 39.09 g/mol -----> m = 0.010 / 39.09 = 2.56x10-4 mol/kg

Ca = 32 mg/kg or 0.032 g/kg; MM = 40.07 g/mol -----> m = 0.032 / 40.07 = 7.99x10-4 mol/kg

Mg = 27 mg/kg or 0.027 g/kg; MM = 24 g/mol -----> m = 0.027 / 24 = 1.125x10-3 mol/kg

SO4 = 280 mg/kg or 0.28 g/kg; MM = 32 + 64 = 96 g/mol ---> m = 0.28 / 96 = 2.92x10-3 mol/kg

Sr = 0.4 mg/kg or 0.0004 g/kg; MM = 87.62 g/mol ----> m = 0.0004 / 87.62 = 4.57x10-6 mol/kg

Cl = 15 mg/kg or 0.015 g/kg; MM = 35.5 g/mol ----> m = 0.015 / 35.5 = 4.23x10-4 mol/kg

HCO3 = 150 mg/kg or 0.15 g/kg; MM = 1+12+48 = 61 g/mol ----> m = 0.15 / 61 = 2.46x10-3 mol/kg

SiO2 = 19 mg/kg or 0.019 g/kg; MM = 32 + 28.08 = 60.08 g/mol -----> m = 0.019 / 60.08 = 3.16x10-4 mol/kg

[H] = antlog (-7.40) = 3.98x10-8 M

This could be considered a salt water basically because of the components of this water. It has too many ions and elements that compounds a salty water. The fresh water has already passed through a process that eliminates the majority of these elements, and only has a few of them in low concentrations.

Answer to b) and c) i'll upload in a few hours. Then, l'll get back and edit my answer.

EDIT:

b) First let's calculate the number of moles of positive charges:

K+, Na+, Ca2+, Mg2+, Sr2+

N1 = 6.52x10-3 + 2.56x10-4 + (7.99x10-4 + 1.125x10-3 + 4.57x10-6)x2 = 0.0106 eq/kg

Number of moles of negative charges:

SO42-, Cl-, HCO3-

N2 = (2.92x10-3)x2 + 4.23x10-4 + 2.46x10-3 = 8.723x10-3 eq/kg

Nt = 0.0106 - 8.723x10-3 = 1.877x10-3 eq/kg

According to this result, there is excess in positive charge and the solution it's not charged balanced.

c) % = 1,877x10-3 / (8.723x10-3 + 0.0106) x 100 = 9.71%

This is a small amount of imbalanced charge actually. This happens because this imbalance make some solutions more salty than others, or even, more acidic or basic.

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