What volume of 0.867 M solution is required to make 3.50 L of a solution with pH of 12.5?
It depends upon the kind of soln doesn't it?
i don't believe so. i've never done a problem like this and i'm not even sure where to start
To determine the volume of the 0.867 M solution needed to make 3.50 L of a solution with a pH of 12.5, we need to follow a few steps:
Step 1: Identify the concentration of H+ ions (or acidity) corresponding to a pH of 12.5. The pH scale is logarithmic, so we need to convert the pH to the concentration of H+ ions using the formula: [H+] = 10^(-pH).
In this case, we have: [H+] = 10^(-12.5) = 3.16 x 10^(-13) M.
Step 2: Recognize that pH is related to the pOH of a solution. In aqueous solutions, pH + pOH = 14. Therefore, pOH = 14 - 12.5 = 1.5.
Step 3: Determine the concentration of OH- ions corresponding to a pOH of 1.5. Using the same logic as in Step 1, we can find [OH-] = 10^(-pOH) = 3.16 x 10^(-2) M.
Step 4: Recognize that in a neutral solution, the concentration of H+ ions and OH- ions are equal, which means [H+] = [OH-].
Step 5: Calculate the volume of the 0.867 M solution needed using the equation: (concentration)(volume) = moles.
Let's assume the volume of the 0.867 M solution needed is V liters.
From Step 1, we know that (3.16 x 10^(-13) M)(3.50 L) = (0.867 M)(V).
Simplifying the equation, we find V = (3.16 x 10^(-13) M)(3.50 L) / (0.867 M).
Now, substitute the values into the equation:
V = (3.16 x 10^(-13) M)(3.50 L) / (0.867 M) ≈ 1.27 x 10^(-10) L.
Therefore, approximately 1.27 x 10^(-10) liters or 0.127 nanoliters of the 0.867 M solution is required to make 3.50 L of a solution with a pH of 12.5.