Jawabanmu

2014-03-28T20:46:13+07:00
[H^+]= \frac{(V HCl. M HCl)-(V NaOH . M NaOH)}{V HCl + V NaOH}
[H^+]= \frac{(10mL . 0,1M)-(9,5mL . 0,1M)}{10 + 9,5}
[H^+]=\frac{1,95}{19,5} = 0,1 = 1. 10^-^1

pH= -log [H^+]
pH= -log 1 . 10^-^1
pH = 1 - log 1
pH = 1 - 0
pH = 1
1 5 1