Consider a solution containing 0.150 M Ba(OH)2. Calulate the pH.
Ba(OH)2 ==> Ba^2+ + 2OH^-
0.150M Ba(OH)2 = 2*0.150M in OH.
pOH = -log(H^+), then pH from
pH + pOH = pKw = 14.0
To calculate the pH of a solution, we need to determine the concentration of hydroxide ions ([OH-]) and then convert it to pOH. Finally, we can use the equation pH = 14 - pOH to find the pH.
In this case, we have a solution containing 0.150 M Ba(OH)2. Since each molecule of Ba(OH)2 dissociates into two hydroxide ions (OH-), we can say that the concentration of hydroxide ions is twice the concentration of Ba(OH)2. Therefore, the [OH-] in the solution is 2 * 0.150 M = 0.300 M.
Now, let's calculate the pOH. The pOH is the negative logarithm (base 10) of the hydroxide ion concentration:
pOH = -log([OH-])
= -log(0.300)
= 0.5229
Finally, we can use the relationship pH = 14 - pOH to find the pH:
pH = 14 - 0.5229
≈ 13.477
Therefore, the pH of the solution containing 0.150 M Ba(OH)2 is approximately 13.477.