# Question b33d8

Jul 27, 2017

I can't give you a "correct" answer with the information given, but I can show you how to solve it, which I did.

#### Explanation:

This is a really annoying unit conversion problem before we can even solve it, hold on:

$\frac{0.4 m o l}{{\mathrm{dm}}^{3}} \cdot \frac{{\mathrm{dm}}^{3}}{c {m}^{3}} \cdot \frac{c {m}^{3}}{m L} \cdot \frac{{10}^{3} m L}{L} = 400 M$

3 NaOH (aq) + C_6H_8O_7 (aq) → Na_3C_6H_5O_7 (aq) + 3 H_2O(l)#

You usually need the volume of one or the other solution in order to accurately solve this problem, but I will assume we are trying to neutralize $1 L$ of citric acid.

$1 L \cdot \frac{0.005 m o l}{L} \cdot \frac{3 N a O H}{{C}_{6} {H}_{8} {O}_{7}} \cdot \frac{L}{400 m o l} \approx 0.375 m L$

This appears to be a bit unreasonable, so perhaps in your problem the amount of citric acid needing to be neutralized is larger.

Jul 28, 2017

The required volume is ${\text{38 cm}}^{3}$.

#### Explanation:

Step 1. Write the balanced equation for the reaction

$\text{H"_3"C"_6"H"_5"O"_7 + "3NaOH" → "Na"_3"C"_6"H"_5"O"_7 + 3"H"_2"O}$

or

$\text{H"_3"Cit" + "3NaOH" → "Na"_3"Cit" + 3"H"_2"O}$

Citric acid is a tribasic acid. Its titration curve will look something like the one below.

Step 2. Calculate the moles of $\text{NaOH}$

$\text{Moles of NaOH" = 0.005 color(red)(cancel(color(black)("mol H"_3"Cit"))) × "3 mol NaOH"/(1 color(red)(cancel(color(black)("mol H"_3"Cit")))) = "0.015 mol NaOH}$

Step 3. Calculate the volume of $\text{NaOH}$

$\text{Volume of NaOH" = 0.015 color(red)(cancel(color(black)("mol NaOH"))) × ("1000 cm"^3 "NaOH")/ (0.4 color(red)(cancel(color(black)("mol NaOH")))) ="38 cm"^3 "NaOH}$

Note: The answer can have only one significant figure, because that is all you gave for the concentration of the $\text{NaOH}$ and for the moles of citric acid.

However, I calculated the answer to two significant figures.