KOH + HCOOH ---> HCOOK + H2O
0,0036 0,003 mol
Theo p.ứ trên, nKOH dư = 0,0006 mol ---> pOH = -log[OH-] = -log(0,0006) = 3,22 ---> pH = 14 - pOH = 10,78.
KOh + Hcl --> KCl + H2O
0.0003 <-0.0003 nKOH du = 0.00006 mol pOH = -log ( 0.0006 / 0.05)= 2.92 => pH = 14-2.92 = 11.08
sr anh nham ty HCOOH + KOH --> HCOOK + H2O
0.003 --> 0.003 => n Koh du =0.00006 =>pOH = -log ( 0.00006/ 0.05) = 2.92
=> ph =11.08