Question

Question 27 5 pts How many grams of lithium fluoride, LiF (MW = 25.94 g/mol), are required for a 65.0 mL of a 0.725 MHF solution to make a buffer at a pH = 3.85? (Ka = 7.2 x 10-4) O 6.23 O 0.380 O 0.247 O 0.329 0 0.0562

Answer 1

det grams of di F required = ng Ka of HF = 7.2010 pka = -log (ka) 2 - log (7:2x104) - = 3.14 Using Henderson Hassebalch equilion pH = pka + log [ salt] [and] det volume of buffer so formed be v We know that, c = m ( number of mole) Concentration V (Total volume C.V=m 25.94 Number of moles of salt = x Number of moles of and HF = ( Molarity) (Volume) Molaily of HFin betti (0-225) (65x10>) com Molarity of salt = 4 x 1

ge Paga No pH = pka 1(25.94) (0725)/65 x 103 3.85 = 3.14 t log [ (2594110713/65 x103 J Solving above equation we get n = 6.239