Calculate the pH of a 0.10 M solution of barium hydroxide, Ba(OH)2.
Ba(OH)2 ==> Ba^ + 2OH^-
So 0.1 M Ba(OH)2 gives you twice that for OH^- in solution or (OH^-) = 0.2 M.
pOH = ??, then pH = 14-pOH.
To calculate the pH of a solution of barium hydroxide (Ba(OH)2), we need to consider the dissociation of the compound in water. Ba(OH)2 is a strong base and dissociates completely in water to form barium ions (Ba2+) and hydroxide ions (OH-).
The dissociation equation for Ba(OH)2 can be written as follows:
Ba(OH)2 -> Ba2+ + 2OH-
Since Ba(OH)2 dissociates to form two hydroxide ions for every one barium ion, the concentration of hydroxide ions in the solution is twice the concentration of barium hydroxide. In other words, [OH-] = 2 × [Ba(OH)2].
Given that the concentration of the barium hydroxide solution is 0.10 M, we can substitute this value into the equation to determine the concentration of hydroxide ions. [OH-] = 2 × 0.10 M = 0.20 M.
Now, to calculate the pOH, we need to take the negative logarithm (base 10) of the hydroxide ion concentration: pOH = -log[OH-].
pOH = -log(0.20) = 0.70.
Finally, to find the pH, we can use the fact that pH + pOH = 14. Rearranging the equation, pH = 14 - pOH.
pH = 14 - 0.70 = 13.30.
Therefore, the pH of a 0.10 M solution of barium hydroxide is 13.30.