number 17 please WL, llllllllllll. after pH of the solutin the 16. An aqueous so

Question

number 17 please

WL, llllllllllll. after pH of the solutin the 16. An aqueous solution of 0.057 M weak acid. Hx, has a pH of 46s. Wha und in the the pH of the solution ifo.018 molof Kxis dissolved in one liter ofthe weak acid 17. Which of the following would form a buffer if added to 2500 mld 0.150 M SnF2? (a) 0.100 mol of HC (b) 0.060 mol of HCI tions of (c) 0.040 mol of HCI (d) 0.040 mol of NaoH (e) 0.040 mol of HF 18. Which of the following would form a buffer if added to 650.0 ml of 0.40 M Sr(OH)2? (a) 1.00 mol of HF (b) 0.75 mol of HF ons of

Explanation / Answer

Answer: 0.100 mol of HCl

Explanation:

2HCl + SnF2 ==> 2HF + SnCl2

This reaction should take place since a weaker acid is formed (HF is weaker than HCl). If there is any F- left over fropm SnF2, then you would have a buffer system (a weak acid, HF, plus its conjugate base, F-).

moles SnF2 = M SnF2 x L SnF2 = (0.150)(0.2500) = 0.0375 moles SnF2
moles HCl = 0.100

0.0375 moles SnF2 x (2 moles HCl / 1 mole SnF2) = 0.0750 moles HCl to completely react with the SnF2. Since we have more than that (0.100 moles HCl), there will be no SnF2 left, and no F- left to form a buffer with the HF produced.

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