$C{{O}_{2}}+2KOH\to {{K}_{2}}C{{O}_{3}}+{{H}_{2}}O$   (1)

0,075$\leftarrow $ 0,15                    (mol)

$C{{O}_{2}}+2NaOH\to N{{a}_{2}}C{{O}_{3}}+{{H}_{2}}O$         (2)

0,125$\to $ 0,25                               (mol)

$\begin{align}& {{n}_{C{{O}_{2}}}}=\frac{{{V}_{C{{O}_{2}}}}}{22,4}=\frac{4,48}{22,4}=0,2\left( mol \right) \\& {{n}_{NaOH}}={{V}_{NaOH}}.{{C}_{{{M}_{NaOH}}}}=0,2.a(mol) \\& {{n}_{KOH}}={{V}_{KOH}}.{{C}_{{{M}_{KOH}}}}=0,2.0,75=0,15\left( mol \right) \\\end{align}$

Theo phương trình (1) có: $\begin{align}  & {{n}_{C{{O}_{2}}pu}}=\frac{{{n}_{KOH}}}{2}=\frac{0,15}{2}=0,075\left( mol \right) \\& \Rightarrow {{n}_{C{{O}_{2}}conlai}}={{n}_{C{{O}_{2}}}}-{{n}_{C{{O}_{2}}pu}}=0,2-0,075=0,125\left( mol \right) \\\end{align}$

Theo phương trình (2) có: ${{n}_{NaOH}}=2.{{n}_{C{{O}_{2}}conlai}}=2.0,125=0,25\left( mol \right)$

Mà $\begin{align}& {{n}_{NaOH}}=0,2.a \\& \Rightarrow 0,2.a=0,25\Rightarrow a=1,25\left( M \right) \\\end{align}$

$\Rightarrow $Chọn đáp án C