Thallium hydroxide is a strong base used in the synthesis of some organic compounds. Calculate the pH of a solution containing 2.55 g TIOH per liter.
Sorry! I'm having a really hard time with this acid/base stuff!
TlOH is a strong base as the problem says. Do it just as you would a NaOH problem. Determine mols from the 2.55 g mass. That will be the mols/L and that is the (OH^-). Then pOH = -log(OH^-) and pH + pOH = 14 to solve for pH. Post your work if you get stuck.
No problem! I can help you step by step with this problem.
1. First, let's find the number of moles of TlOH in the solution.
To do this, we'll use the molar mass of TlOH. The molar mass of TlOH can be calculated by summing the atomic masses of the elements present in it:
Tl (thallium) = 204.3833 g/mol
O (oxygen) = 15.9994 g/mol
H (hydrogen) = 1.00784 g/mol
So, the molar mass of TlOH = 204.3833 + 15.9994 + 1.00784 = 221.39054 g/mol
Now, we can find the number of moles by dividing the mass (2.55 g) by the molar mass:
Moles of TlOH = 2.55 g / 221.39054 g/mol
2. Next, we need to calculate the concentration of hydroxide ions (OH-) in the solution.
Since TlOH is a strong base, it dissociates completely in water, producing one hydroxide ion (OH-) for every molecule of TlOH.
Therefore, the concentration of hydroxide ions (OH-) is equal to the molar concentration of TlOH. In this case, it is the same as the number of moles calculated in step 1 as we are assuming a 1:1 ratio between TlOH and OH-.
Concentration of OH- = Moles of TlOH / Volume (in liters)
3. Now, let's calculate the pOH. The pOH is calculated using the formula:
pOH = -log[OH-]
4. Finally, we can calculate the pH using the relationship:
pH + pOH = 14
Just substitute the calculated pOH value into this equation to get the pH.
Remember, if you encounter any difficulties during these steps, feel free to ask for further clarification!