Question

Sebanyak 100 mL asam lemah HCN 0.1 M yang mempunyai PH 3 di campur dgn 100 mL NaOH 0.1 M. Hitunglah PH larutan yg terbentuk

Answer 1

8+log2,2 ada pilihan jawabannya gaa??

Answer 2

PH = 3
pH = - log [H+]
3 = - log [H+]
[H+] = 10⁻³
[H+] = √ Ka x M
10⁻³ = √Ka x 0,1
10⁻⁶ = Ka x 0,1
Ka = 10⁻⁵

mmol HCN = 100 x 0,1 = 10 mmol
mmol NaOH = 100 x 0,1 = 10 mmol
     HCN + NaOH => NaCN + H2O
a : 10          10
b : 10          10            10        10
s :  -            -              10        10

[NaCN] = 10 mmol / 200 ml = 0,05 M
[OH-] = √Kw/Ka x M
[OH-] = √10⁻¹⁴/10⁻⁵ x 0,05
[OH-] = √ 5 x10⁻¹¹
[OH-] = 7,07 x10⁻⁶

pOH = - log[OH-]
pOH = - log 7,07 x10⁻⁶
pOH = 6 - log 7,07

pH = 14 - pOH
pH = 14 - (6 - log 7,07)
pH = 8 + log 7,07
pH = 8 + 0,85 = 8,85