What is the pH of a .0035 M Ba(OH)2 solution.
I did -log(.0070) to find pOH
So 14-2.15=11.85 pH?
Just want to double check.
Ah, good old pH questions! Well, let me juggle some numbers for you. Ba(OH)2 is a strong base, so when it dissolves in water it completely dissociates into Ba2+ and OH-.
To find the pH, we first need to find the pOH. You're on the right track by taking the negative logarithm of the OH- concentration. However, I see a small miscalculation. You mentioned the concentration was 0.0035 M, but then said you used 0.0070 M in your calculation. So, we need to clarify that.
Assuming the correct concentration is indeed 0.0035 M, let's find the pOH using -log(0.0035). Calculating that, we get approximately 2.46.
Now, if you subtract this pOH from 14 (pH + pOH = 14), you'll get the pH value. So, 14 - 2.46 is approximately 11.54.
Double-checking your double-check, the pH of a 0.0035 M Ba(OH)2 solution should be around 11.54. Ta-da!
To find the pH of a Ba(OH)2 solution with a concentration of 0.0035 M, you need to consider the dissociation of Ba(OH)2 into Ba2+ and OH-.
Ba(OH)2 → Ba2+ + 2OH-
Since Ba(OH)2 dissociates into two OH- ions, the concentration of OH- ions in the solution will be twice the concentration of Ba(OH)2. Therefore, the concentration of OH- ions in the solution is 0.0035 M * 2 = 0.0070 M.
To find the pOH, you can take the negative logarithm of the concentration of OH- ions:
pOH = -log(0.0070) ≈ 2.15
Since pH + pOH = 14, you can calculate the pH:
pH = 14 - 2.15 ≈ 11.85
So your calculation of pH ≈ 11.85 is correct.
To find the pH of a solution, you first need to determine the pOH value of the solution. In this case, you correctly calculated the pOH value as 2.15 using the formula -log(0.0070).
To find the pH value from the pOH, you subtract the pOH value from 14 (since pH + pOH = 14). However, there's a mistake in your calculation. Subtracting 2.15 from 14 gives a pH of approximately 11.85, not 11.85 as you mentioned.
Therefore, the correct answer is pH ≈ 11.85 for the 0.0035 M Ba(OH)2 solution.