# Question 41ff8

Sep 22, 2016

Here's what I got.

#### Explanation:

As it turns out, the values given to you by the problem are inaccurate. Here's why I say that.

Start by examining the balanced chemical equation that describes this neutralization reaction -- I won't add the sodium cations because they are spectator ions

${\text{C"_ 6"H"_ 8"O"_ (7(aq)) + color(blue)(3)"OH"_ ((aq))^(-) -> "C"_ 6 "H"_ 5"O"_ (7(aq))^(3-) + 3"H"_ 2"O}}_{\left(l\right)}$

Notice that it takes $\textcolor{b l u e}{3}$ moles of hydroxide anions to neutralize $1$ mole of citric acid, which basically means that the reaction will always consume three times more moles of hydroxide anions than moles of citric acid.

Use the molarity and volume of the sodium hydroxide solution to find how many moles of hydroxide anions, which are delivered in a $1 : 1$ mole ratio by sodium hydroxide, are present in the sample

6.89 color(red)(cancel(color(black)("cm"^3))) * (1color(red)(cancel(color(black)("dm"^3))))/(10^3color(red)(cancel(color(black)("cm"^3)))) * "1 mole OH"^(-)/(1color(red)(cancel(color(black)("dm"^3))))

$= {\text{0.00689 moles OH}}^{-}$

This many moles of hydroxide anions were needed to neutralize

0.00689 color(red)(cancel(color(black)("moles OH"^(-)))) * ("1 mole C"_6"H"_8"O"_7)/(color(blue)(3)color(red)(cancel(color(black)("moles OH"^(-)))))

$= {\text{0.002297 moles C"_6"H"_8"O}}_{7}$

Now, you know that you have $0.002297$ moles of citric acid in ${\text{10 cm}}^{3}$ of solution. This implies that the ${\text{250 cm}}^{3}$ solution you made by dissolving the hydrate contained

250 color(red)(cancel(color(black)("cm"^3))) * ("0.002297 moles C"_6"H"_8"O"_7)/(10color(red)(cancel(color(black)("cm"^3))))

$= {\text{0.05743 moles C"_6"H"_8"O}}_{7}$

Keep in mind that these are moles of anhydrous citric acid. Use the molar mass of anhydrous citric acid to calculate the mass you had in the $\text{10-g}$ hydrate sample

0.05743 color(red)(cancel(color(black)("moles C"_6"H"_8"O"_7))) * "192.124 g"/(1color(red)(cancel(color(black)("mole C"_6"H"_8"O"_7))))#

$= \text{ 11.03 g}$

As you can see, this is more than the $\text{10-g}$ sample of the hydrate you used to make the first solution.

The mass of the hydrate used to make the ${\text{250 cm}}^{3}$ solution must be greater than $\text{11.03 g}$ because it contains a certain mass of water of crystallization.

Once you get the correct value for the mass of the hydrate, simply

• calculate the mass of the water of crystallization by subtracting the mass of the anhydrous citric acid, i.e. $\text{11.03 g}$, from the total mass of the hydrate
• use the molar mass of water to convert that mass to moles