Question

Please answer all the questions

#1. HA is a weak acid. an aqueous solution of HA was prepared by dissolving 0.185 mol of HA and diluted to 525mL. The pH of solution was 3.93 at 25.0°C. The Ka of HA is _____.

#2. What is the pH of a solution at 25°C with[H^+] = 2.50 • 10^-5?

Answer 1

1. 
   HA Copy+ A caq, HA is a weak and weak and does not conize completely cag) + H₂O (d) 2 Hot Ka = [H264] [Alex +7 [A] Ka Kao ionization constant / Dissociation constant All concentrations are all equilibriums Initially HA is 0.185 mot in 525 ml (0.5256) molarity - no.of mol (mel 0. 185 mol Volume (4) = 0.352 mol/L=0.352M 0.525L Now, we need to find out equilibrium concentrations using ICE table. [HAJeg Hot - X tx ka = x2 HA A Initial 0.352 M Change +x Equilibrium (6-352m.n) (26) (0) (2) (x) 0.352M-X 0.352M-X but here, ple is given as = 3.93 and pea-dog [H₃O+] So, 3.93 = -log[Hz0] - [H30+) = 16 = 0.0001174M ,:. [H30+] (H30+] - [A-] = 0.0001174 Then ka (0.0001174) 2 1.38 x 10 (0.352-0.0001174) 0.35798 ka -3.93 -8 کم 3.92 X 10 8 TH+] = 2.50 X10; pH !? pH og [Ht -log (2.5x105) 4.6
   thanks and hope you like the answer