Đề bài chắc là tính pH dung dịch thu được ha.
\(n_{OH^-}=0.25\cdot0.004+0.25\cdot0.003=0.00175\left(mol\right)\)
\(n_{H^+}=0.25\cdot0.0035+0.25\cdot2\cdot0.004=0.002875\left(mol\right)\)
\(OH^-+H^+\rightarrow H_2O\)
\(0.00175..0.00175\)
\(n_{H^+\left(dư\right)}=0.002875-0.00175=0.001125\left(mol\right)\)
\(V_{dd}=250+250=500\left(ml\right)=0.5\left(l\right)\)
\(\left[H^+\right]=\dfrac{0.001125}{0.5}=0.00225\left(M\right)\)
\(pH=-log\left[H^+\right]=-log\left(0.00225\right)=2.64\)