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

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

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Answer 1 · 2017-05-17T16:05:04

You need 40 g of $NaOH$ $"NaOH"$ .

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 $C_{23} H_{16} O_{6}$ $"C"_23"H"_16"O"_6$ .

We can write the formula as $H_{2} C_{23} H_{14} O_{6}$ $"H"_2"C"_23"H"_14"O"_6$ .

Then the equation for the neutralization is

$M_{r} : m 338.375 m m m 39.997$ $M_text(r):color(white)(m)338.375color(white)(mmm)39.997$
$m m H_{2} C_{23} H_{14} O_{6} + 2NaOH \to {Na}_{2} C_{23} H_{14} O_{6} + 2 H_{2} O$ $color(white)(mm)"H"_2"C"_23"H"_14"O"_6 + "2NaOH" → "Na"_2"C"_23"H"_14"O"_6 + 2"H"_2"O"$
$m m m m l H_{2} A m l l + 2NaOH \to m m {Na}_{2} A m l l + 2 H_{2} O$ $color(white)(mmmml)"H"_2"A"color(white)(mll) + "2NaOH" → color(white)(mm)"Na"_2"A" color(white)(mll)+ 2"H"_2"O"$

Step 1. Calculate the moles of $H_{2} A$ $"H"_2"A"$

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

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 $"NaOH"$

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

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

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