\(n_{NaOH}=0.05\cdot0.4=0.02\left(mol\right)\)
\(n_{HCl}=0.05\cdot0.2=0.01\left(mol\right)\)
\(NaOH+HCl\rightarrow NaCl+H_2O\)
Ta có :
\(\dfrac{n_{NaOH}}{1}>\dfrac{n_{HCl}}{1}\) \(\Rightarrow NaOHdư\)
\(n_{NaOH\left(dư\right)}=0.02-0.01=0.01\left(mol\right)\)
\(\left[OH^-\right]=\dfrac{0.01}{0.05+0.05}=0.1\)
\(pH=14+log\left(0.1\right)=13\)
nH+=0.5x0.04=0.02(mol)
nOH-=0.5x0.06=0.03(mol)
=>nOH- dư, tính theo số mol H+
OH-+H+-->H2O
0.02 0.02 0.02 (mol)
nOH- dư =0.03-0.02=0.01(mol)
CMOH-=0.01/(0.04+0.06)=0.1(M)
=>pOH-=-log(0.1)=1
=>pH=14-1=13