IN THE REACTION BETWEEN SODIUM HYDROXIDE AND TETRAOXOSULPHATE (VI) SOLUTIONS, WHAT VOLUME OF 0.5 MOLAR SODIUM HYDROXIDE WOULD EXACTLY NEUTRALIZE 10CM3 OF 1.25 MOLAR TETRAOXOSULPHATE (VI) ACID

The reaction between sodium hydroxide (NaOH) and tetraoxosulphate (VI) solutions can be represented by the following balanced chemical equation:

H2SO4 + 2NaOH → Na2SO4 + 2H2O

From the balanced equation, it is clear that 1 mole of sulfuric acid (H2SO4) reacts with 2 moles of sodium hydroxide (NaOH). Therefore, the molar ratio of H2SO4 to NaOH is 1:2.

Given that the concentration of the tetraoxosulphate (VI) acid is 1.25 M and the volume is 10 cm3, we can calculate the number of moles of H2SO4 present in the solution using the formula:

moles = concentration × volume

moles = 1.25 mol/L × 0.01 L moles = 0.0125 moles

According to the molar ratio, the number of moles of NaOH required to neutralize the H2SO4 is twice that of H2SO4. Therefore, the number of moles of NaOH required is:

moles of NaOH = 2 × moles of H2SO4

moles of NaOH = 2 × 0.0125

moles of NaOH = 0.025 moles

Now, we can use the formula for calculating the volume of a solution given its concentration and number of moles:

volume = moles / concentration

volume = 0.025 moles / 0.5 mol/L

 volume = 0.05 L volume = 50 cm3

Therefore, the volume of 0.5 molar sodium hydroxide required to exactly neutralize 10 cm3 of 1.25 molar tetraoxosulphate (VI) acid is 50 cm3.