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# pH and [OH-] of Basic Solution

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[OH-] and the pH for each of the following strong base solutions:

-- 0.082 M KOH

-- 1.065 g of KOH in 500.0 mL of solution

-- 10.0 mL of 0.0105 M Ca(OH)2 diluted to 500.0 mL

-- a solution formed by mixing 10.0 mL of 0.015 M Ba(OH)2 with 40.0 mL of 7.5x10-3 M NaOH.

https://brainmass.com/chemistry/physical-chemistry/acid-base-equilibria-concentrations-ph-214935

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Acid-Base Equilibria
I am unsure of how to calculate

[OH-] and the pH for each of the following strong base solutions:
The pOH is calculated as pOH = - log[OH-]
And pOH + pH = 14

-- 0.082 M KOH
In solution, KOH is a strong base and electrolyte, so
KOH (aq) → K+ (aq) + OH- (aq)
0.82 0.082
Thus, the concentration of OH- is
[OH-] = 0.082
Then
pOH = - log[OH-] = - log0.082 = 1.086
pH = 14 - pOH = 14 - 1.086 = 12.914

-- 1.065 g of KOH in 500.0 mL of solution
the moles of KOH is 1.065 g / 56 g/mol = 0.019 mol
So the concentration of 500.0 mL = 0.5 L solution is
0.019 mol / 0.5 L = 0.038 M
So the concentration of OH- is
[OH-] = 0.038 M
Then
pOH = - log[OH-] = - log0.038 = 1.42
pH = 14 - pOH = 14 - 1.42 = 12.58

-- 10.0 mL of 0.0105 M Ca(OH)2 diluted to 500.0 mL
Ca(OH)2 (aq) → Ca2+ (aq) + 2OH- (aq)
Then [OH-] = 2 * 0.0105 = 0.021 M in the 10.0 mL solution.
The mole of OH- is 10.0 * 0.021 = 0.21 mmol
After the dilution, the concentration is then
0.021 / 500.0 = 0.00042 M
[OH-] = 0.00042 M
pH = 14 + log[OH-] = 10.6

-- a solution formed by mixing 10.0 mL of 0.015 M Ba(OH)2 with 40.0 mL of 7.5x10-3 M NaOH
Ba(OH)2 (aq) → Ba2+ (aq) + 2OH- (aq)
The concentration of OH- in this solution itself is 0.015 M * 2
The moles of OH- from Ba(OH)2 is
10.0 * 0.015 * 2 = 0.3 mmol

The moles of OH- from NaOH is
40.0 * 0.0075 = 0.3 mmol
So the total OH- is 0.6 mmol
The volume of the solution is 10.0 + 40.0 = 50.0 mL
So
[OH-] = 0.6 mmol / 50 mL = 0.012 M
pH = 14 + log[OH-] = 12.08.

This content was COPIED from BrainMass.com - View the original, and get the already-completed solution here!