Question

0,64 gram asam iodine / Hl (Mr= 128) dilarutkan dalam air sampai volume 500 ml, berapa pH larutan???

Answer 1

M = gr/mr x 1000/v
      0,64/128 x 1000/500
      = 0.01
H+ = x. ma
        1. 0.01
       = 0.01
        = 1x10-2
PH = log H+ 
        2 log 1