posted by john .
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.