This document contains worked examples for an environmental engineering homework assignment on acid-base chemistry calculations. It includes calculating molarity, normality, and concentrations from other units. Example problems cover solubility products, pH calculations, alkalinity, and converting between ion concentrations and nitrogen concentrations. Step-by-step workings are shown for each multi-part question, demonstrating how to set up and solve the relevant chemical equations to obtain the requested values.

1. Answer to
CE 441 Environmental Engineering
Homework #2
(Due 09/19/2013)
1. Calculate the molarity and normality of (a) 100 mg/L of H2SO4 and (b) 100
mg/L of HCl.
Molarity = concentration in g/L divided by molecular weight in g/mole. If
concentration is in other units, you will need to convert to g/L. That way you will get
the correct unit for molarity, which is mole/L or M.
Normality is molarity multiplied by number of charges.
Answer:
Molarity: (a) 1.02 X10^-3M; Nomality: 2.04 X10^-3N
(b) 2.74 X10^-3M; Nomality: 2.74 X10^-3N
2. Find mg/L of (a) 0.02 N Ca++ and (b) 0.5 M HCO3-.
For this question, you reverse the above process if normality is used. For Ca++,
0.02N is equivalent to 0.01M or mole/L. Multiplied by the molecular weight, you will
get concentration in g/L. However since the question asks for results in the unit of
mg/L, you will need to convert g/L to mg/L by multiplying 1000.
Answer: (a): 400; (b): 30,500
3. The solubility product of calcium fluoride (CaF2) is 3.45 x 10-11 M3. Will a
fluoride concentration of 1.0 mg/L be soluble in a water sample containing
200 mg/L of calcium?
You need to find out the molar concentrations of Ca++ and F-, then calculate the ion
product of [Ca++]*[F-]2 and compare it with Ksp. If it’s larger than Ksp, some of
the ions will precipitate since Ksp is the upper limit of the ion product that can be
soluble in solution.
Answer: Yes (you need to state the reason why)
4. If 10 mg of H2SO4 is added to 1L of water, what is the final pH? What
amount of NaOH, in mg, would be required to neutralize this acid solution?

2. A reaction equation needs to be established first. In this case, it is:
H2SO4 == 2H+ + SO42So molar ratio between H2SO4 and 2H+ is 1:2, or one mole of H2SO4 produces 2
moles of H+. By converting 10mg of H2SO4 to molar amount, [H+] can be calculated,
then pH can be calculated by –log[H+].
Answer: final pH = 3.69; 8.16mg of NaOH
5. Calculate the pH of a water sample at 25 oC that contains 0.62 mg/L of
carbonic acid. Assume that [H+] = [HCO3-] at equilibrium and neglect
dissociation of water.
The equation needs to be established first:
H2CO3 == H+ + HCO3There are 2 ways of understanding the problem. I had assumed that the amount of
H2CO3 was the INITIAL amount, and it will change after it dissociates into H+ and
HCO3-. Some students assumed that this 0.62 mg/L was the final amount after the
dissociation happens and an equilibrium is established – both can be correct, but the
latter is easier to solve than the first scenario.
Regardless, you will need to convert 0.62 mg/L to molar concentration first for
H2CO3, which is 1x10-5M (refer to question 1 for procedure).
You also need to know what Ka1 stands for, and the governing equation for Ka1,
which is:
Ka1 = 10-6.35 = [H+][HCO3-]/[H2CO3]
First scenario:
After dissociation completes and an equilibrium is established, if we assume x moles
of H2CO3 dissociates, the molar amounts for H2CO3, H+, and HCO3- will be:
1x10-5-x, x, and x, respectively.
Bring these amounts to the Ka1 equation above, where x is the only unknown, and
you will get an answer by solving a quadratic equation. If you establish the equation
correct and the calculation afterwards is wrong, you still get 75% credit.
Second scenario:
The molar amounts for H2CO3, H+, and HCO3- will be: 1x10-5, x, and x,
respectively. Following the same procedure, solving x becomes a simple square root
calculation, and you will get a slightly different answer.

3. Answer: pH=5.73 (scenario 1); 5.67 (scenario 2)
6. Convert the following from mg/L as the ion to mg/L as CaCO3:
(a) 100 mg/L Ca++
(b) 100 mg/L CO2
(c) 100 mg/L HCO3(d) 100 mg/L CO3=
The equation for this is
mg/L as the ion to mg/L as CaCO3 = mg/L of ion * EW of CaCO3/EW of ion.
EW of CaCO3 is always 1000/2 = 50;
EW for other ions is molecular weight divided by charge.
The tricky part for CO2 is that it should be treated as H2CO3, which carries 2
charges, so EW (CO2) = 44/2 = 22.
Answer: (a) 250 (b) 227; (c) 82.0; (d) 167
7. Calculate the “approximate” alkalinity (in mg/L as CaCO3) of water
containing 120 mg/L of bicarbonate ion and 10 mg/L of carbonate ion.
Alkalinity is calculated by:
Alk = [CO32-] + [HCO3-] + [OH-] + [H+]
In this case, [OH-] and [H+] are both negligible ([CO32-] and [HCO3-] coexist only
in nearly neutral water, hence both [OH-] and [H+] are very small).
Therefore, the equation becomes:
Alk = [CO32-] + [HCO3-]
Follow the procedure in question 1 and 6, the two terms on the right can be easily
calculated.
Answer: 115
8. Convert (a) 20 mg/L NO3- and (b) 20 mg/L NH3 in mg-N/L.

4. The calculation involves finding molar amount of chemical species and do the
subsequent conversion. For example, the first question, not all these 20mg/L of NO3
is N – much of it is oxygen. In fact, for NO3, percentage of nitrogen is MWN/MWNO3,
or 14/62. This is the conversion factor you need. Similarly, the conversion factor for
NH3 is 14/17.
Answer: (a) 4.52; (b) 16.5