\(n_{NaOH}=0,25.V\left(mol\right)\)
\(n_{Ca\left(OH\right)_2}=0,5.V\left(mol\right)\)
=> \(n_{OH^-}=0,25.V+2.0,5.V=1,25V\left(mol\right)\)
\(n_{HCl}=0,55.2=1,1\left(mol\right)=>n_{H^+}=1,1\left(mol\right)\)
H+ + OH- --> H2O
1,1->1,1
=> 1,25.V = 1,1
=> V = 0,88(l)