Question

1.berapa gram massa larutan KOH yg harus disediakan,jika kita inginmembuat 250 larutan tsb dengan ph=11? (Ar K=39 ; O=16 ; H=1)

2.HITUNGLAH PH BERIKUT
a.0,001 mol HNO3 100 ml
b.0.1 gram NaOH yg ditambah air hingga 250 ml
c.0,01 mol CH3COOH dlm 100 ml (Ka=1*10-5
d.0,05 mol NH3 dlm 500 ml larutan(Kb=1*10-5)

Answer 1

1. 250 ml KOH, Basa kuat
    pH = 11 ---> pOH = 3  ----> [OH⁻] = 0,001 M
    [KOH] = (g/Mr) x (1000/v)
     0,001 = (g/56) x 1000/250)
     g = 0,056/4= 0,014 gram
2. a. [HNO₃] = 0,001 mol/0,1L = 0,01 M
       pH = - log 0,01 = 2
    b. [NaOH] = (1/40)x (1000/250) = 0,1 M
       pOH = - log 0,1 = 1 -----> pH = 13
    c. [CH₃COOH] = 0,01 mol/0,1L = 0,1 M
        [H⁺] = √10⁻⁵x10⁻¹ = 10⁻³ M
        pH = - log 10⁻³ = 3
    d. [NH₃] = 0,05 mol/0,5L = 0,1 M
        [OH⁻] = √10⁻⁵x10⁻¹ = 10⁻³ M
         pOH = - log 10⁻³ = 3 -----> pH = 11