0.233 g of barium sulfate can be obtained from a 10 cm3 solution of sulfuric acid. What volume of NaOH

0.233 g of barium sulfate can be obtained from a 10 cm3 solution of sulfuric acid. What volume of NaOH solution (w = 8%, p = 1.09) is needed to completely neutralize 75 cm3 of such an acid solution.

Given: V1 solution (H2SO4) = 10 cm ^ 3 m (BaSO4) = 0.233 g ω (NaOH) = 8% ρ solution (NaOH) = 1.09 g / cm ^ 3 V2 solution (H2SO4 ) = 75 cm ^ 3
Find: V solution (NaOH) -?
Decision:
1) Write the reaction equation: 2NaOH + H2SO4 => Na2SO4 + 2H2O; 2) Find the amount of substance BaSO4: n (BaSO4) = m (BaSO4) / Mr (BaSO4) = 0.233 / 233 = 0.001 mol; 3) Find the amount of substance S in BaSO4: n (S) = n (BaSO4) = 0.001 mol; 4) Find the amount of substance H2SO4 in a solution of 10 cm ^ 3: n1 (H2SO4) = n (S) = 0.001 mol; 5) Find the amount of H2SO4 in a 75 cm ^ 3 solution: Create a proportion and solve: if 10 cm ^ 3 of an H2SO4 solution contains 0.001 mol of H2SO4, then 75 cm ^ 3 contains X mol; 10 – 0.001; 75 – x; x = 75 * 0.001 / 10 = 0.0075; n2 (H2SO4) = x = 0.0075 mol; 6) Find the amount of NaOH substance (taking into account the reaction equation): n (NaOH) = n2 (H2SO4) * 2 = 0.0075 * 2 = 0.015 mol; 7) Find the mass of the NaOH substance: m (NaOH) = n (NaOH) * Mr (NaOH) = 0.015 * 40 = 0.6 g; 8) Find the mass of the NaOH solution: m solution (NaOH) = m (NaOH) * 100% / ω (NaOH) = 0.6 * 100% / 8% = 7.5 g; 9) Find the volume of NaOH solution: V solution (NaOH) = m solution (NaOH) / ρ solution (NaOH) = 7.5 / 1.09 = 6.88 cm ^ 3.
Answer: The required volume of NaOH solution is 6.88 cm ^ 3.