\(n_{H_2SO_4}=\dfrac{2,45}{98}=0,025\left(mol\right)\Rightarrow n_{H^+}=0,025\cdot2=0,05\left(mol\right)\)
\(n_{NaOH}=0,3\cdot0,15=0,045=n_{OH^-}\)
PT ion RG: \(H^++OH^-\rightarrow H_2O\)
0,045 ← 0,045
\(\Rightarrow n_{H^+\text{ (dư)}}=0,05-0,045=0,005\left(mol\right)\)
\(\Rightarrow\left[H^+\right]_{\text{(dư)}}=\dfrac{0,005}{0,35+0,15}=0,01\left(M\right) \)
\(\Rightarrow pH=-\log0,01=2\)