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san4es73 [151]
3 years ago
5

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

cipitate all but 0.3 mg/L of the iron? The temperature of the solution is 25˚C.
Chemistry
1 answer:
xenn [34]3 years ago
4 0

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

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