\(\left[H^+\right]=0,2\cdot0,01=0,002mol\\ \left[OH^-\right]=0,3\cdot0,002=0,0006mol\\ H^++OH^-\rightarrow H_2O\)
0,02 > 0,0006
\(n_{H^+dư}=0,002-0,0006=0,0014mol\\ \)
\(\Sigma_{dd}=\dfrac{0,0014}{0,5}=0,0028M\\ \Rightarrow pH\approx2,6\)
nHCl=0,01. 0,2=2.10-3 nNaOH=0,002.0,3=6.10-4
HCl --> H+ + Cl- NaOH --> Na+ + OH -
2.10-3-->2.10-3 6.10-4 --> 6.10-4
H+ + OH - --> H2O (dư H+)
6.10-4 6.10-4
=> nH+ = 1,4 .10-3
[H+ ] = (1,4.10-3) / 0,5=2,8.10-3
==> pH= -Lg(2,8.10-3) =2,55
