Question

Use the Henderson-Hasslebalch equation to calculate the pH of each solution:

(Express your answer using two decimal places)

Part A) A solution that contains 0.620% C5H5N by mass and 0.900% C5H5NHCl by mass (Kb for C5H5N is 1.7 x 10^-9)

Part B) A solution that is 16.5 g of HF and 27.0 g of NaF in 125 mL of solution (Ka for HF is 3.5x10^-4)

Answer 1

A) pKof CH N =-log K = -log(1.7x10 ")= 8.77 PK, of C,H-N = 14.00-pky = 14.00 -8.77 = 5.23 - 0.620x10 Concentration of C,H, N, (CHN)= 9 - = 0.0784M 79.1 Concentration of CH NH*, [CH wt fu wot_0.900x10 -= 0.0778M 15.6 [C,H,N] According to Henderson equation, pH = pK, +10g [C,H,NH'] 0.0784 = 5.23 +10g 0,0778 = 5.23 - - - - - - - - - -------------- 165. 1 mol 16.5 g *20.08. - 6.60M B)Concentration of HF (HF= 0125 L 1 mol 27.08 *4208.5.14 M Concentration of F, F 0.125 L PK, of HF = -log K = -log(3.5x10+) = 3.46 According to Henderson equation F pH = PK, +logie = 3.46+ 10g 6.60 = 3.35