\(n_{CuSO_4}=1\cdot0,3=0,3\left(mol\right);n_{NaOH}=2\cdot0,07=0,14\left(mol\right)\\ PTHH:CuSO_4+2NaOH\rightarrow Na_2SO_4+Cu\left(OH\right)_2\downarrow\\ \text{Vì }\dfrac{n_{CuSO_4}}{1}>\dfrac{n_{NaOH}}{2}\Rightarrow CuSO_4\text{ dư}\\ \Rightarrow n_{Na_2SO_4}=\dfrac{1}{2}n_{NaOH}=0,07\left(mol\right)\\ \Rightarrow C_{M_{Na_2SO_4}}=\dfrac{0,07}{0,3+0,07}\approx1,19M\)