Question

A chemist uses a solution of 0.226 M (KOH) to titrate 89.07 mL (H2SO4), she finds that it requires 34.00 mL of the base to reach the end-point of the titration. What is the molarity of the acid solution ? must be rounded to the correct number of sig figs

Answer 1

`"0.0431 M"`.

Explanation:

The question basically tells us that `"34.00 mL"` of a `"0.226 M"` solution of `KOH` neutralised `"89.07 mL"` of an unknown concentration of `H_2SO_4`, assuming that end point equals neutralisation point.

Here's the balanced equation of the neutralisation reaction between `KOH` and `H_2SO_4`:

`2KOH + H_2SO_4 -> 2H_2O + K_2SO_4`

From this, we can tell that, for every `2` moles of `KOH` that reacts, `1` mole of `H_2SO_4` will react.
In other words, the number of moles of `H_2SO_4` will be half the number of moles of `KOH`.

Then, we need to find the number of `KOH` that reacted using this equation:

`"number of moles" = "concentration (M)" xx "volume (L)"`
`"number of moles" = "0.226 M" xx "34.00 mL"`
`"number of moles" = "0.226 M" xx "0.03400 L" = "0.00768 moles"`

Since the number of moles of `H_2SO_4` will be half the number of moles of `KOH`, it will be:

`("0.00768 moles " KOH)/2 = "0.00384 moles " H_2SO_4`

So, `"0.00384 moles"` will correspond to `"89.07 mL"` (which was given in the question), we can now find the concentration of `H_2SO_4`:

`"concentration (M)" = "number of moles"/"volume (L)"`

`"concentration (M)" = "0.00384 moles"/"89.07 mL"`

`"concentration (M)" = "0.00384 moles"/"0.08907 L" = "0.0431 M"`