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?

2 Answers
Mar 27, 2018

[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

Mar 28, 2018

"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"