How many grams of 5.400 wt% aqueous HF are required to provide a 50% excess to react with 14.0 mL of 0.0236 M Th4+ by the following reaction?
Th4+ + 4F - → ThF4 (s)
moles Th^4+ = M x L = 0.0236M x 0.014L = 0.0003304.
Convert to moles HF.
0.0003304 x 4 = 0.001322.
Convert to grams.
g = moles x molar mass = 0.001322 x approximately 20 = about 00264 g HF.
Add 50% excess which is approximately 0.00264 + 0.00132 = about 0.04 g.
Your solution of HF is 5.4 wt% which means 5.4g HF/100 g soln. How many g of the solution will give you about 0.04 g HF? I assume you can take it from here. You need to go through the above calculations for I have estimated here and there.
To solve this problem, we need to follow these steps:
Step 1: Determine the number of moles of Th4+ in 14.0 mL of 0.0236 M Th4+ solution.
We'll use the formula: moles = concentration × volume
moles of Th4+ = 0.0236 M × 0.0140 L
Step 2: Use the stoichiometry of the reaction to find the number of moles of HF required.
From the balanced equation, we can see that 1 mole of Th4+ reacts with 4 moles of F-. Therefore, moles of F- = 4 × moles of Th4+
Step 3: Convert the number of moles of F- to grams of aqueous HF.
We need to find the mass of HF in grams. To do this, we'll use the molar mass of HF and the mole-to-mass conversion:
mass of HF = moles of F- × molar mass of HF
Step 4: Calculate the grams of 5.400 wt% aqueous HF required.
To do this, we need to know the wt% (weight percent) of HF in the aqueous solution. Let's assume the wt% means that 5.400 g of HF is present in 100 g of the aqueous solution. This means that the solution contains 5.400 g of HF per 100 g of solution.
We have already calculated the mass of HF in grams from Step 3. We'll use this mass and the ratio of grams of HF to grams of the aqueous solution to find the mass of the aqueous solution required.
mass of aqueous HF = mass of HF / (5.400 g HF/100 g solution)
Now, let's calculate the solution step by step.
Step 1:
moles of Th4+ = 0.0236 M × 0.0140 L = 0.0003296 moles Th4+
Step 2:
moles of F- = 4 × moles of Th4+ = 4 × 0.0003296 moles of F- = 0.0013184 moles F-
Step 3:
molar mass of HF = 1.01 g/mol + 19.00 g/mol = 20.01 g/mol
mass of HF = moles of F- × molar mass of HF = 0.0013184 moles F- × 20.01 g/mol = 0.026387 g HF
Step 4:
mass of aqueous HF = mass of HF / (5.400 g HF/100 g solution)
mass of aqueous HF = 0.026387 g HF / (5.400 g HF/100 g solution) = 0.488 g aqueous HF
Therefore, approximately 0.488 grams of 5.400 wt% aqueous HF are required to provide a 50% excess to react with 14.0 mL of 0.0236 M Th4+.