Question

A 42.0 mL sample of a 0.309 M aqueous hydrofluoric acid solution is titrated with a 0.340 M aqueous barium hydroxide solution. What is the pH after 11.8 mL of base have been added? pH =

Answer 1

Number of milimales of HE acid in 42mL of solution = 42m2 x o: 309m = 0.0424x0.309 molt = 0.01297 mole = 12.97 mmol Number of milimole of Ba(OH) buse in 11.8 m2 of solution = 11.8 m2 x 0.34M = 0.01184*0.34 moll = 0.00401 mol =) 4.01 mmol Number of milimole of HE aid left in the solution aften addition of Ba(OH), base bese = {12.97-4.01) mmol =( = 8.96 mmol Total volume of the solution = 42m2 + 11.8 m2 = 53.8m2 8.96 mmol of HF is acid in 53-8 me of the solution. Molarity of the HF in 53.8 ml af solution 8.96 mmol 538 mal I cromat 8.46X16 mol 0.0538 8.96X163 m 53.88103 = 0.166 mol 2.0 166 mol of HF (ht) present in the solution

8Ł:0 btŁ.0. = (991:0) for- = Hd [A] Bop - = Hd