Question

What is the strength in g/ litre of a solution of `H_2SO_4`,12 ml of which neutralized 15 ml of N/10 NaOH solution?

Answer 1

The solution contains 6.1 g of `"H"_2"SO"_4` per litre of solution.

Explanation:

Step 1. Write the balanced equation

`"2NaOH + H"_2"SO"_4 → "Na"_2"SO"_4 + 2"H"_2"O"`

Step 2. Calculate the equivalents of `"NaOH"`

`"Equivalents" = 0.015 color(red)(cancel(color(black)("L NaOH"))) × "0.1 eq NaOH"/(1 color(red)(cancel(color(black)("L NaOH")))) = "0.0015 eq NaOH"`

Step 3. Calculate the equivalents of `"H"_2"SO"_4`

In a reaction, 1 eq of anything is equivalent to 1 eq of anything else.

Hence, there are 0.0015 eq of `"H"_2"SO"_4`.

Step 4. Calculate the moles of `"H"_2"SO"_4` in the sample

`"Mass of H"_2"SO"_4 = 0.0015 color(red)(cancel(color(black)("eq H"_2"SO"_4))) × (1 color(red)(cancel(color(black)("mol H"_2"SO"_4))))/(2 color(red)(cancel(color(black)("eq H"_2"SO"_4)))) = "0.000 75 mol H"_2"SO"_4`

Step 5. Calculate the mass of `"H"_2"SO"_4` in the sample

`"Mass of H"_2"SO"_4 = "0.000 75" color(red)(cancel(color(black)("mol H"_2"SO"_4))) × ("98.08 g H"_2"SO"_4)/(1 color(red)(cancel(color(black)("mol H"_2"SO"_4)))) = "0.074 g H"_2"SO"_4`

Step 6. Calculate the mass of `"H"_2"SO"_4` in 1 L of solution

`"Mass of H"_2"SO"_4 = 1000 color(red)(cancel(color(black)("mL solution"))) × ("0.074 g H"_2"SO"_4)/(12 color(red)(cancel(color(black)("mL solution")))) = "6.1 g H"_2"SO"_4`