Question

In a titration experiment, a student finds that 25.49 mL of an NaOH solution is needed to neutralize 0.7137 g of a potassium hydrogen phthalate (KHP, a monoprotic acid with the formula KHC8H4O4). What is the concentration of the NaOH solution?

Answer 1

`[NaOH]=0.1371*mol*^-1`

Explanation:

We (i) write out the equation we investigate...`"KHP"` is a monoprotic acid...`"KHP"-=1,2-C_6H_4(CO_2H)(CO_2^(-)K^(+))`

`1,2-C_6H_4(CO_2H)(CO_2^(-)K^(+))+NaOH(aq) rarrC_6H_4(CO_2^(-)Na^(+))(CO_2^(-)K^(+))+H_2O(l)`

And (ii) we assess the molar quantity of `"KHP"`...

`"Moles of KHP"=(0.7137)/(204.22 *g·mol^-1)=0.00349476*mol`

And because `"KHP"` here acts as a mono-acid...this is the molar quantity of `NaOH`...which were delivered in a volume of `25.49*mL`.

And so `[NaOH]=(0.00349476*mol)/(25.49*mLxx10^-3*L*mL^-1)`

`[NaOH]=0.1371*mol*^-1`

Answer 2

`"0.1371M"`.

Explanation:

This is the formula for molarity:

`"molarity of NaOH" = "number of moles of NaOH"/"volume of NaOH solution (L)"`

First, we need to find the number of moles of `NaOH`:

`KHP` being "monoprotic" means that one mole of `KHP` is one equivalent. Basically, `1` molecule of `KHP` only donates `1` `H^+` ion.

We also know that `NaOH` is a monoprotic base, because there's only one `OH^-` ion in its chemical formula.

Therefore, `1` mole of `KHP` will correspond to `1` mole of `NaOH` in a neutralisation reaction.
In other words, `1` mole of `NaOH` will neutralise `1` mole of `KHP`.

The number of moles of `KHP` that was neutralised was:

`"0.7137 g" / (39.098 + 1.008 + 8 xx 12.01 + 4 xx 1.008 + 4 xx "16.00 g/mol")`

`= 0.003495` moles

Because the mole ratio of `KHP` to `NaOH` is `1:1`, `0.003495` moles of `NaOH` must have neutralised `0.003495` moles of `KHP`.

Then, we need to find the volume of the `NaOH` solution. This is pretty simple, actually, because it was given to us in the question: `"25.49 mL"`, or `"0.02549 L"`.

Finally, we just need to plug these values into the formula for molarity:

`"molarity of NaOH" = "number of moles of NaOH"/"volume of NaOH solution (L)"`

`"molarity of NaOH" = "0.003495 mol"/"0.02549 L"`

`"molarity of NaOH = 0.1371 M"`