ANSWER
The molarity of NaOH is 0.933 M
EXPLANATION
Given that;
The volume of NaOH is 30.00mL
The volume of H2SO4 is 35.00mL
The molarity of H2SO4 is 0.400 M
The number of moles of the acid (nA) = 1
The number of moles of the base(nB) = 2
The balanced equation of the reaction is
[tex]\text{ H}_2SO_4\text{ + 2NaOH }\rightarrow\text{ Na}_2SO_4\text{ + 2H}_2O[/tex]Follow the steps below to find the molarity of NaOH
Step 1; Apply the dilution formula
[tex]\text{ }\frac{\text{ C}_{A\text{ }}V_A}{\text{ C}_B\text{ V}_B}\text{ = }\frac{\text{ n}_A}{\text{ n}_B}[/tex][tex]\begin{gathered} \text{ }\frac{0.4\text{ }\times\text{ 35}}{\text{ C}_B\times\text{ 30}}\text{ = }\frac{\text{ 1}}{\text{ 2}} \\ \text{ cross multiply} \\ \text{ 0.4 }\times\text{ 35 }\times\text{ 2 = C}_B\times\text{ 30 }\times\text{ 1} \\ \text{ 28 = 30C}_B \\ \text{ Divide both sides by 30} \\ \text{ C}_{B\text{ }}\text{ = }\frac{\text{ 28}}{\text{ 30}} \\ \text{ C}_{B\text{ }}\text{ = 0.933 M} \end{gathered}[/tex]Therefore, the molarity of NaOH is 0.933 M
