What is the pOH of a 0.032 M Ca(OH)2 (strong base) solution?
i solved it by doing
14=-log (0.064)
14-1.19= 12.81
but the right answer is 1.19, why? i thought the OH in the problem meant i had to use 14?
You calculated the pH not the pOH
pOH=-log([OH]/mole l^-1)
pH=-log([H]/mole l^-1)
Note that pOH=14-pH is only valid at 25C.
To calculate the pOH of a solution, you need to use the concentration of hydroxide ions (OH-) in the solution, not the concentration of the base itself. In the case of a strong base like Ca(OH)2, when it dissolves in water, it dissociates completely to yield two moles of OH- ions for every mole of Ca(OH)2.
Since the concentration of Ca(OH)2 is given as 0.032 M, the concentration of OH- ions is twice that: 0.032 M × 2 = 0.064 M.
The pOH is defined as the negative logarithm (base 10) of the hydroxide ion concentration.
So, pOH = -log[OH-] = -log(0.064) ≈ 1.19
Therefore, the correct pOH value for a 0.032 M Ca(OH)2 solution is indeed 1.19.
To find the pOH of a solution, you need to first determine the concentration of hydroxide ions (OH-) in the solution. In the case of a strong base like Ca(OH)2, each formula unit of Ca(OH)2 dissociates into two OH- ions.
Given the concentration of Ca(OH)2 as 0.032 M, the concentration of OH- ions will be twice that value, i.e., 2 × 0.032 M = 0.064 M.
pOH is defined as the negative logarithm (base 10) of the hydroxide ion concentration. So, to find the pOH, you would take the negative logarithm of 0.064:
pOH = -log (0.064) ≈ 1.19 (rounded to two decimal places)
The pOH scale represents the concentration of hydroxide ions in a solution, just like the pH scale represents the concentration of hydrogen ions. The pH scale ranges from 0 to 14, with 0 being highly acidic and 14 being highly basic. Therefore, when finding the pOH, you do not need to assume a value of 14. Instead, you calculate the pOH directly using the formula mentioned above. In this case, the correct answer is indeed 1.19.