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# pH Adjustment of a Aolution with NaOH or Ca(OH)2

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Assume you have a 64000L volume of solution (predominantly water) with a pH of 4.03.

1. How much NaOH would be required to raise the pH to 5.5?

2. Or, if using Ca(OH)2 --- how much calcium hydroxide would be required to raise the pH of this solution to 5.5?

In each case, assume full ionization of the base.

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#### Solution Preview

1) pH = 5.5 means pOH = 14 - 5.5 = 8.5
hence, for pOH = 5.5,
hence, using relation, pOH = -log [OH-]
or [OH-] = log-1 (-pOH) = log-1 (-8.5) = 3.16 x 10^-9 M
therefore, concentration of OH-Ions ( or NaOH) = 3.16 x 10^-9 M.

Now, for pH = 4.03, pOH = 14 - 4.03 = 9.97
hence, [OH-] = log-1(-pOH) = log-1(-9.97) = 1.07x 10^-10 M
since, concentration = moles/volume,
so ...

#### Solution Summary

This solution calculates how much sodium hydroxide and calcium hydroxide is required to raise the pH from 4.03 to 5.5 by using the power of hydrogen equation. All steps are shown with brief explanations.

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