Question

calcule el ph de una disolucion 0,24 M sabiendo que k=1,8× 10^-5
NH3------> NH+4+OH-​

Answer 1

NH3   →   NH4+    +    OH-

Co - x          x                  x

K = x²/Co - x

K = x²/0,24 - x

1,8 x 10^-5 = x²/0,24 - x

0 = x² + 1,8 x 10^-5x - 4,32 x10^-6

x = 2,069 x 10^-3

pOH = -log(2,069x10^-3)

pOH = 2,68

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 2,68

pH = 11,32

SALUDOS!