Calculate the pOH and pH at 298K of solutions having the following ion concentrations
A.[OH-] = 1.0 X 10^-12
B.[OH-] = 1.3 X 10^-2
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To calculate the pOH and pH, we can use the equations:
pOH = -log[OH-]
pH = 14 - pOH
Let's calculate the pOH and pH for each scenario:
A. [OH-] = 1.0 x 10^-12
pOH = -log[1.0 x 10^-12]
pOH = -(-12) = 12
pH = 14 - pOH
pH = 14 - 12 = 2
Therefore, for a solution with [OH-] = 1.0 x 10^-12, the pOH is 12 and the pH is 2.
B. [OH-] = 1.3 x 10^-2
pOH = -log[1.3 x 10^-2]
pOH = -(-1.88) = 1.88
pH = 14 - pOH
pH = 14 - 1.88 = 12.12
Therefore, for a solution with [OH-] = 1.3 x 10^-2, the pOH is 1.88 and the pH is 12.12.
To calculate the pOH and pH of a solution, we need to know the concentration of hydrogen ions (H+) or hydroxide ions (OH-) in the solution. In this case, we are given the concentration of hydroxide ions ([OH-]).
To calculate the pOH, we can use the formula:
pOH = -log[OH-]
Let's calculate the pOH for each case:
A. [OH-] = 1.0 X 10^-12
Using the formula, we have:
pOH = -log(1.0 X 10^-12)
= -log(10^-12)
= -(-12) log(10)
= 12
So, the pOH of solution A is 12.
To calculate the pH, we can use the formula:
pH = 14 - pOH
For case A, we have:
pH = 14 - 12
= 2
So, the pH of solution A is 2.
B. [OH-] = 1.3 X 10^-2
Using the formula, we have:
pOH = -log(1.3 X 10^-2)
= -log(0.013)
= -log(1.3) - log(10^-2)
= -0.113 - (-2)
= 2.113
So, the pOH of solution B is approximately 2.113.
To calculate the pH for case B, we can again use the formula:
pH = 14 - pOH
For case B, we have:
pH = 14 - 2.113
= 11.887
So, the pH of solution B is approximately 11.887.
Therefore, the pOH and pH for the given ion concentrations at 298K are:
A. [OH-] = 1.0 X 10^-12
pOH = 12
pH = 2
B. [OH-] = 1.3 X 10^-2
pOH ≈ 2.113
pH ≈ 11.887