a)
\(n_{HCl}=\frac{6,72}{22,4}=0,3\left(mol\right)\)
=> CM = \(\frac{0,3}{3}=0,1M\)
=> pH = -log(0,1) = 1
b)
\(n_{H^+}=0,3\left(mol\right)\)
Có \(pH=14+log\left(OH^-\right)\)
=> \(log\left(OH^-\right)=-1\)
=> \(C_{M\left(OH^-\right)}\) (sau pư) = 0,1M
=> \(C_{M\left(NaOH\right)}\) (sau pư) = 0,1M
=> \(n_{NaOH\left(saupu\right)}=0,1.\left(3+2\right)=0,5\left(mol\right)\)
PTPL: \(H^++OH^-\rightarrow H_2O\)
______0,3--->0,3______________(mol)
=> \(n_{NaOH\left(pư\right)}=0,3\left(mol\right)\)
=> CM (dd NaOH ban đầu) = \(\frac{0,3+0,5}{2}=0,4M\)