Thallium(I) hydroxide is a strong base used in the synthesis of some organic compounds. Calculate the pH of a solution containing 2.57 g TlOH per liter.
I got 1.44 but I'm pretty sure that's wrong?
But that's close for the pOH.
mols TlOH = grams/molar mass = ?= mols OH^-
pOH = -log (OH^-)
Then pH + pOH = pKw = 14. You know pK2 and pOH, solve for pH.
To calculate the pH of a solution containing Thallium(I) hydroxide (TlOH), we need to determine the concentration of hydroxide ions (OH-) in the solution.
First, let's calculate the number of moles of TlOH in 2.57 g. We can use the molar mass of TlOH, which is 221.36 g/mol.
Number of moles = mass / molar mass
Number of moles = 2.57 g / 221.36 g/mol
Number of moles ≈ 0.0116 mol
Since TlOH is a strong base, it fully dissociates in water, meaning that one mole of TlOH will produce one mole of hydroxide ions (OH-).
Therefore, the concentration of OH- ions in the solution is also 0.0116 M.
Now, to calculate the pH, we need to determine the pOH first. The pOH is the negative logarithm of the OH- concentration.
pOH = -log[OH-]
pOH = -log(0.0116)
pOH ≈ 1.93
Finally, we can calculate the pH using the equation: pH = 14 - pOH.
pH = 14 - 1.93
pH ≈ 12.07
So, the pH of the solution containing 2.57 g TlOH per liter is approximately 12.07, not 1.44.