Prompt

Groundwater in Pherric, New Mexico, initially contains 1.800 mg/L of iron as Fe3+. What must the pH be raised to in order to pre

xenn [34]

Answer : The pH will be, 3.2

Explanation :

As we known that the value of solubility constant of ferric hydroxide at $25^oC$ is, $2.79\times 10^{-39}$

Amount or solubility of iron consumed = (1.800 - 0.3) mg/L = 1.5 mg/L

The given solubility of iron convert from mg/L to mol/L.

$1.5mg/L=\frac{1.5\times 10^{-3}g/L}{56g/mol}=2.7\times 10^{-7}mol/L$

The chemical reaction will be:

$Fe(OH)_3\rightarrow Fe^{3+}+3OH^-$

The expression of solubility constant will be:

$K_{sp}=[Fe^{3+}]\times [3OH^-]^3$

Now put all the given values in this expression, we get the concentration of hydroxide ion.

$2.79\times 10^{-39}=(2.7\times 10^{-7})\times [3OH^-]^3$

$[OH^-]=1.5\times 10^{-11}M$

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (1.5\times 10^{-11})$

$pOH=10.8$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-10.8\\\\pH=3.2$

Therefore, the pH will be, 3.2

4 0

3 years ago

Given the ion C2O4-2, what species would you expect to form with each of the following ions?

Ksivusya [100]

Answer:

A. K₂C₂O₄ Potassium oxalate

B. CuC₂O₄ Copper oxalate

C. Bi₂(C₂O₄)₃ Bismuth (III) oxalate

D. Pb(C₂O₄)₂ Lead (IV) oxalate

E. (NH₄)₂C₂O₄ Ammonium oxalate

F. HC₂O₄⁻ Acid oxalate

Explanation:

C₂O₄⁻² → oxalate anion

This is the conjugate base from the H₂C₂O₄ which is the oxalic acid. A weak dyprotic acid that can release 2 protons.

A. 2K⁺ + C₂O₄⁻² → K₂C₂O₄ Potassium oxalate

It can be formed by the neutralization of the acid with the base

H₂C₂O₄ + 2KOH → K₂C₂O₄ + 2H₂O

B. Cu²⁺ + C₂O₄⁻² ⇄ CuC₂O₄ ↓

This is a precipitate.

C. 2Bi³⁺ + 3C₂O₄⁻² ⇄ Bi₂(C₂O₄)₃ ↓

This is a precipitate.

D. Pb⁴⁺ + 2C₂O₄⁻² ⇄ Pb(C₂O₄)₂ ↓

This is a precipitate.

E. 2NH₄⁺ + C₂O₄⁻² ⇄ (NH₄)₂C₂O₄ ↓

This is a precipitate.

F. This is the conjugate strong base, for the weak acid

H⁺ + C₂O₄⁻² ⇄ HC₂O₄⁻

HC₂O₄⁻ + H₂O ⇄ C₂O₄⁻² + H₃O⁺ Ka

HC₂O₄⁻ + H₂O ⇄ H₂C₂O₄ + OH⁻ Kb

HC₂O₄⁻ is an amphoteric compound

6 0

3 years ago

Write the nuclear symbol of oxygen-16?

Nina [5.8K]

Answer:

Oxygen-16 (16o) is a stable isotope of oxygen, having 8 neutrons and 8 protons in its nucleus.

Explanation:

5 0

3 years ago

What percentage of radioactive substance remains after two half-lives

Ksju [112]

Answer:

After 2 half-lives there will be 25% of the original isotope, and 75% of the decay product. After 3 half-lives there will be 12.5% of the original isotope, and 87.5% of the decay product. After 4 half-lives there will be 6.25% of the original isotope, and 93.75% of the decay product.

Explanation:

3 0

3 years ago

C3H8(g) + 5O2(g) ⟶ 3CO2(g) + 4H2O(g) H = -2220 kJ If 865.9 g of H2O is produced during this combustion, how much heat is generat

dem82 [27]

Answer:

3 × 10⁴ kJ

Explanation:

Step 1: Write the balanced thermochemical equation

C₃H₈(g) + 5 O₂(g) ⟶ 3 CO₂(g) + 4 H₂O(g) ΔH = -2220 kJ

Step 2: Calculate the moles corresponding to 865.9 g of H₂O

The molar mass of H₂O is 18.02 g/mol.

865.9 g × 1 mol/18.02 g = 48.05 mol

Step 3: Calculate the heat produced when 48.05 moles of H₂O are produced

According to the thermochemical equation, 2220 kJ of heat are evolved when 4 moles of H₂O are produced.

48.05 mol × 2220 kJ/4 mol = 2.667 × 10⁴ kJ ≈ 3 × 10⁴ kJ

5 0

2 years ago