Question

A 0.110 M `H_2SO_4` solution is used to neutralize 10.0 mL of 0.085 M `NaOH`. What volume of the acid, in mL, is required to neutralize the base?

Answer 1

You need 3.86 mL of `"H"_2"SO"_4` to neutralise the base.

Explanation:

So what we have to do is use stoichiometry to find the volume of acid needed to neutralise the base.

1: We need a balanced chemical equation:

`"H"_2"SO"_4(aq) + "2NaOH (aq)" rarr "Na"_2"SO"_4 (s) + "2H"_2"O"(l)`

So what we can find first is the moles of `"NaOH"`.

2: Solve for moles. Note that I changed 10 mL to 0.01 L.

`n_1 = cv`

`= "0.085 mol/L (0.010 L)"`

`= 8.5xx10^-4 "mol"`

3: Now we transfer the moles of `"NaOH"` to `"H"_2"SO"_4"`.

To do that, we divide by `"NaOH"`'s coefficient, and multiply by `"H"_2"SO"_4`'s.

`n_2 = (8.5xx10^-4 "mol")/2`

`= 4.25xx10^-4 "mol"`

There are `4.25xx10^-4` mol of `"H"_2"SO"_4`.

4: With `"H"_2"SO"_4`'s moles and concentration (given), we can solve for volume.

`v = n_"2"/c`

`= (4.25xx10^-4 "mol")/("0.110 mol/L")`

`"= 0.003,863,636"` L = `"3.863,636"` mL

Significant digits tells us to round the volume to 3 decimal places:

`3.86` mL.

Therefore, you need `3.86` mL of `"H"_2"SO"_4` to neutralise the base.

Hope this helps :)