Question

Trộn 50,0ml dd NaOH 0,40M với 50,0 ml dd HCl 0,20M được dd A. Tính pH của dd A

 

Answer 1

\(n_{NaOH}=0.05\cdot0.4=0.02\left(mol\right)\)

\(n_{HCl}=0.05\cdot0.2=0.01\left(mol\right)\)

\(NaOH+HCl\rightarrow NaCl+H_2O\)

Ta có : 

\(\dfrac{n_{NaOH}}{1}>\dfrac{n_{HCl}}{1}\) \(\Rightarrow NaOHdư\)

\(n_{NaOH\left(dư\right)}=0.02-0.01=0.01\left(mol\right)\)

\(\left[OH^-\right]=\dfrac{0.01}{0.05+0.05}=0.1\)

\(pH=14+log\left(0.1\right)=13\)

Answer 2

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)

CM_OH-=0.01/(0.04+0.06)=0.1(M)

=>pOH-=-log(0.1)=1

=>pH=14-1=13