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

Question

Avery Adams · Accepted Answer

You need 40 g of #"NaOH"#.

Assuming that pamoic acid is that soluble, here's how you would do the calculation. Pamoic acid is a dibasic acid with the formula #"C"_23"H"_16"O"_6#. We can write the formula as #"H"_2"C"_23"H"_14"O"_6#. Then the equation for the neutralization is #M_text(r):color(white)(m)338.375color(white)(mmm)39.997# #color(white)(mm)"H"_2"C"_23"H"_14"O"_6 + "2NaOH" → "Na"_2"C"_23"H"_14"O"_6 + 2"H"_2"O"# #color(white)(mmmml)"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 #"H"_2"A"# #"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 #"NaOH"# #"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 #"NaOH"# #"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"#

Luna Chen · Answer

undefined The molar mass of pamoic acid is 517.6 g/mol. Therefore, you would need 0.5176 moles of pamoic acid to neutralize 500 mL of it. As sodium hydroxide reacts with pamoic acid in a 1:1 ratio, you would need 0.5176 moles of sodium hydroxide. The mass of sodium hydroxide required is 23.0 g.