# What mass of sodium hydroxide do I need to neutralize 500 mL of 1 mol/L pamoic acid?

May 17, 2017

You need 40 g of $\text{NaOH}$.

#### Explanation:

Assuming that pamoic acid is that soluble, here's how you would do the calculation.

Pamoic acid is a dibasic acid with the formula ${\text{C"_23"H"_16"O}}_{6}$.

We can write the formula as ${\text{H"_2"C"_23"H"_14"O}}_{6}$.

Then the equation for the neutralization is

${M}_{\textrm{r}} : \textcolor{w h i t e}{m} 338.375 \textcolor{w h i t e}{m m m} 39.997$
$\textcolor{w h i t e}{m m} \text{H"_2"C"_23"H"_14"O"_6 + "2NaOH" → "Na"_2"C"_23"H"_14"O"_6 + 2"H"_2"O}$
$\textcolor{w h i t e}{m m m m l} \text{H"_2"A"color(white)(mll) + "2NaOH" → color(white)(mm)"Na"_2"A" color(white)(mll)+ 2"H"_2"O}$

Step 1. Calculate the moles of $\text{H"_2"A}$

$\text{Moles of H"_2"A" = 0.500 color(red)(cancel(color(black)("L H"_2"A"))) × ("1 mol H"_2"A")/(1 color(red)(cancel(color(black)("L H"_2"A")))) = "0.50 mol H"_2"A}$

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

$\text{Moles of NaOH" = 0.50 color(red)(cancel(color(black)("mol H"_2"A"))) × "2 mol NaOH"/(1 color(red)(cancel(color(black)("mol H"_2"A")))) = "1.0 mol NaOH}$

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

$\text{Mass of NaOH" = 1.0 color(red)(cancel(color(black)("mol NaOH"))) × "39.997 g NaOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "40 g NaOH}$