Question

What mass of sodium hydroxide will neutralize 60.00 mL of 0.015 mol/L chloric acid?

Answer 1

`"0.036 g NaOH"`

Explanation:

Sodium hydroxide and chloric acid react in a `1:1` mole ratio to produce water and aqueous sodium chlorate, as shown by the chemical equation

`"NaOH"_ ((aq)) + "HClO"_ (3(aq)) -> "H"_ 2"O"_ ((l)) + "NaClO"_ (3(aq))`

This tells you that the reaction will consume equal numbers of moles of sodium hydroxide and of chloric acid.

Use the molarity and volume of the chloric acid solution to calculate how many moles of solute it contained

`60.00 color(red)(cancel(color(black)("mL solution"))) * (1color(red)(cancel(color(black)("L"))))/(10^3color(red)(cancel(color(black)("mL")))) * "0.015 moles HClO"_3/(1color(red)(cancel(color(black)("L solution"))))`

`= 9.00 * 10^(-4)color(white)(.)"moles HClO"_3`

So, if the reaction consumed `9.00 * 10^(-4)` moles of chloric acid, then it must have also consumed `9.00 * 10^(-4)` moles of sodium hydroxide, since the two reactants react in a `1:1` mole ratio.

To convert the number of moles of sodium hydroxide to grams, use the compound's molar mass

`9.00 * 10^(-4) color(red)(cancel(color(black)("moles NaOH"))) * "39.997 g"/(1color(red)(cancel(color(black)("mole NaOH")))) = color(darkgreen)(ul(color(black)("0.036 g")))`

The answer is rounded to two sig figs, the number of significant figures you have for the molarity of the chloric acid solution.

Answer 2

The mass of sodium hydroxide is 36 mg.

Explanation:

The equation for the reaction is

`"HClO"_3 + "NaOH" → "NaClO"_3 + "H"_2"O"`

1. Calculate the moles of `"HClO"_3`.

`"Moles of HClO"_3 = "0.060 00" color(red)(cancel(color(black)("L NaOH"))) × "0.015 mol HClO"_3/(1 color(red)(cancel(color(black)("L HClO"_3)))) = "0.000 900 mol HClO"_3`

2. Calculate the moles of `"NaOH"`.

`"Moles of NaOH" = "0.000 900" color(red)(cancel(color(black)("mol HClO"_3))) × "1 mol NaOH"/(1 color(red)(cancel(color(black)("mol HClO"_3)))) = "0.000 900 mol NaOH"`

3. Calculate the mass of the `"NaOH"`.

`"Mass of NaOH" = "0.000 900" color(red)(cancel(color(black)("mol NaOH"))) × "40.00 g NaOH"/(1 color(red)(cancel(color(black)("mol NaOH")))) = "0.036 g NaOH" = "36 mg NaOH"`