Question

There are copper(III)ions in wastewater that needs to precipitate to copper(II)hydroxide at ph 9.2. Wastewater pH is 3.5. PH is adjusted with 50-w% NaOH. NaOH solution density is 1.43 kg/l. How much you need NaOH when wastewater flow is 3800 m3/d?

And then the second question: How much copperhydroxide dissolves to water (25 celsius) when it´s solubility product is 4.8*10^20?

Answer 1

Part 1. Warning! Long Answer. You will need to use 71 L of the `"NaOH"` solution per day.

Explanation:

1. Calculate the hydronium concentration in the wastewater

`"pH = 3.5"`

`["H"_3"O"^"+"] = 10^"-3.5"color(white)(l)"mol/L" = 3.2 × 10^"-4"color(white)(l)"mol/L"`

2. Calculate the amount of `"H"_3"O"^"+"` in `"3800 m"^3` of wastewater

`"Moles of H"_3"O"^"+"`

`= 3800 color(red)(cancel(color(black)("m"^3color(white)(l) "wastewater"))) × (10^3 color(red)(cancel(color(black)("L wastewater"))))/(1 color(red)(cancel(color(black)("m"^3color(white)(l) "wastewater")))) × (3.2 × 10^"-4"color(white)(l)"mol H"_3"O"^"+")/(1 color(red)(cancel(color(black)("L wastewater")))) = "1200 mol H"_3"O"^"+"`

3. Calculate the moles of `"NaOH"` required to neutralize the acid

The equation for the reaction is

`"NaOH + H"_3"O"^"+" → "Na"^"+" + 2"H"_2"O"`

`"Moles of NaOH" = 1200 color(red)(cancel(color(black)("mol H"_3"O"^"+"))) × "1 mol NaOH"/(1 color(red)(cancel(color(black)("mol H"_3"O"^"+")))) = "1200 mol NaOH"`

4. Calculate the concentration of `"OH"^"-"` at pH 9.2

`"pH = 9.2"`

`"pOH = 14.00 - pH = 14.00 - 9.2 = 4.8"`

`["OH"^"-"] = 10^"-pOH"color(white)(l)"mol/L" = 10^"-4.8"color(white)(l)"mol/L" = 1.6 × 10^"-5" color(white)(l)"mol/L"`

5. Calculate the amount of `"OH"^"-"` in `"3800 m"^3` of wastewater

`"Moles of OH"^"-"`

`= 3800 color(red)(cancel(color(black)("m"^3color(white)(l) "wastewater"))) × (10^3 color(red)(cancel(color(black)("L wastewater"))))/(1 color(red)(cancel(color(black)("m"^3color(white)(l) "wastewater")))) × (1.6 × 10^"-5"color(white)(l)"mol OH"^"-")/(1 color(red)(cancel(color(black)("L wastewater")))) = "60 mol OH"^"-"`

6. Calculate the total moles of `"NaOH"` required

`"Moles of NaOH = (1200 + 60) mol NaOH = 1260 mol NaOH"`

7. Calculate the mass of `"NaOH"`

`"Mass of NaOH solution"`

`= 1260 color(red)(cancel(color(black)("mol NaOH"))) × "40.00 g NaOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "50 500 g NaOH = 50.5 kg NaOH"`

8. Calculate the mass of `"NaOH"` solution

`"Mass of NaOH" = 50.5 color(red)(cancel(color(black)("kg NaOH"))) × "100 kg NaOH solution"/(50 color(red)(cancel(color(black)("kg NaOH")))) = "101 kg NaOH solution"`

9. Calculate the volume of `"NaOH"` solution

`"Volume of NaOH" = 101 color(red)(cancel(color(black)("kg NaOH solution"))) × "1 L NaOHsolution"/(1.43 color(red)(cancel(color(black)("L NaOH solution")))) = "71 L NaOH solution"`

You will need to use 71 L of the `"NaOH"` solution per day.

Answer 2

Part 2. The solubility is `"22 mg·m"^"-3"`.

Explanation:

`color(white)(mmmmml)"Cu(OH)"_2 → "Cu"^"2+" + 2"OH"^"-"`
`"E/mol·L"^"-1": color(white)(mmmmmmm)scolor(white)(mmml)2s`

`K_text(sp) = ["Cu"^"2+"]["OH"^"-"]^2 = s(2s)^2 = 4s^3 = 4.8 ×10^"-20"`

`s^3 = (4.8 × 10^"-20")/4 = 1.2 × 10^"-20"`

`s = 2.3 × 10^"-7"`

`"Solubility" = scolor(white)(l)"mol/L"`

`= 2.3 × 10^"-7"color(red)(cancel(color(black)("mol")))"/L" × "97.56 g"/(1 color(red)(cancel(color(black)("mol")))) = 2.2 × 10^"-5"color(white)(l)"g/L = 0.022 g·m"^"-3" = "22 mg·m"^"-3"`