Question

You may want to reference (Pages 659 - 662) section 16.3 while completing this problem. Calculate the pH of the solution that results from each of the following mixtures. Part A 160.0 mL of 0.25 M HF with 2250 mL of 0.32 M NaF The K, of HF is 6.6 x 10-4 Express your answer using two decimal places. ITO AL OO ? pH = Submit Request Answer Part B

Answer 1

Answer-a 0.32 (M) Given, 160.0mL of is mixed with 225.0mL of 0.25 (M) HF NaF solution. After mixing the total volume of the solution = 160.0+ 225.0 mL = 385.0 mL . Concentration of HF = 160 x 0.25 = 0.1039 (M) 385 385 Concentration of Na F = 225 x 0.32 = 0.1870 (M) we know, pH of buffer solution = pka + log salt acid] 0 • 1870 i pH of the solution = 0.1039 = 3.1804 + 0.255 2 = 3.4356 3.44

Answer-b iven, 170.0 mL of 0-12 (M) C₂ Hg NH₂ is mixed with 285.0 mL of 0.22(M) C₂H₂ NH₃ d. After mixing, total volume of the solution = 17000 + 28500 ml = 455.0 mL i concentration of GHENH = 170 0·12 0·0448 (M) 455 concentration of C HG NHACO - 285x 0.22 0.1378 (M) 455 We know, s pot of buffer = pkb + log I salt to » por of the solution - - log (5-6x104) + log 0.1378 = 3.2518 + 0.4880 = 3.7398 x 3.74 We know, pH + POH = 14 pH of the solution - 14-3.74 = 10.26