Question

Problem 4 (8 points). What is the pH of a 0.890 M NH (aq) solution at room temperature? The base ionization constant of ammonia is Ko = 1.78 10. Round off your answer to 2 decimal places and enter your answer in the box provided with correct units and sig. figs.: The pH of Answer: . He 0.890 M ammonia is): pH =

Answer 1

den NH us a weak base and NH3 + H₂O - 0.890 reaches with water NHT & OH 0 0 Initial Change th tk Equilibrium 0.890-x 0.890 Now K = [NHy] [OH-] [NH₂ xin - 1.70 xlos - w² 0.890 0.890 1 4 + Thus But x = 0.890 x 1.70 x 105 x = 1.5842 x 105 = 15.842 x 10-6 x = √15.842 x166 x = 3.98 x 163 [ou] = x= 3.98 x 103 [H] [OH-] = 1.0 x1074 Ch+] = 10 x 10 3.98 x 103 Ch] = 0.25 xlo" 2.5 x 162

Now Now PH = -log Cut pH = -log (2.5 x1012) a pH = - {log 2.5 + log 1072) pH = -(0.390 -12) [pH = -(-11.602) = +11.6 like if understood, comment for more help.