Question

Question 8 (10 points) Calculate the pH of a solution made by mixing 100.0 mL of 0.644 MNH; with 100.0 ml. of 0.100 MHCI. (K for NH; = 1.8 x 10-) 9.99 4.01 8.52 9.26 none of these Question 9 (10 points) Calculate the pH of a solution that is 0.50 Min HF (K-72 x 10-) and 0.66 Min NaF 3.14 3.26 b BE

Answer 1

question 8 answer is 9.99

NH → loomi, 0-644M HCl → loomi 0.1M number of moles (n = Molarityx Volume (in 1000 NHy moles = (100) (0.644) - = 1000 0.0644 moli HCI moles = (100) (0.1) . 1000 - 0.01 mol The reaction is Nth + Initially, 0.0644 - HCl orol NHACI change: -0.01 -0.01 +0.01 Finally : 0.0544 0.01 Nth and NHACI is a buffer solution. Basic buffer pot = pkb loal number of moles of salt) i buffer pot = pro + log (number of moles of base? But given Kb = 1.8xros CSscanned with CamScanner

pky = -10g Ky = -109 (1.8x10-5) = -109 (1.8) – 10g (1045) = -109 (1.8) -(-5) log do = -0.255 +5 pkb = 4.745 . salt moles vs 0.01, base moles = 0.0544 Then pol = 4.745 +log / 0.0544 = 4.745+ (-0.7356) = 4.009 But we know that pH +POH = 14 Then pH = 14-poh = 14-4:009 pH = 9.99 :: PH = 9.99