Question

How do I find the mass of copper(II) hydroxide which is formed when excess copper(II) sulfate is added to 100mL of a 0.450 mol `L^-1` solution of sodium hydroxide?

Answer 1

`"2.20 g"`

Explanation:

An alternative approach is to simply use the mole ratios that exist between the species that take part in the reaction, then use the molar mass of copper(II) hydroxide, `"Cu"("OH")_2`, to get the mass formed by the reaction.

So, your sodium hydroxide solution contains

`C = n/V implies n = C * V`

`n_(OH^(-)) = "0.450 M" * 100 * 10^(-3)"L" = "0.0450 moles OH"""^(-)`

The net ionic equation for this double replacement reaction looks like this

`"Cu"_text((aq])^(2+) + color(red)(2)"OH"_text((aq])^(-) -> "Cu"("OH")_text(2(s]) darr`

So, `color(red)(2)` moles of hydroxide ions will react with one mole of copper(II) ions and form one mole of copper(II) hydroxide.

This means that the reaction will produce

`0.0450color(red)(cancel(color(black)("moles OH"""^(-)))) * ("1 mole Cu"("OH")_2)/(color(red)(2)color(red)(cancel(color(black)("moles OH"""^(-))))) = "0.0225 moles Cu"("OH")_2`

To find how many grams of copper(II) hydroxide will contain this many moles, use the compound's molar mass

`0.0225color(red)(cancel(color(black)("moles OH"""^(-)))) * "97.56 g"/(1color(red)(cancel(color(black)("mole OH"""^(-))))) = "2.195 g Cu"("OH")_2`

Rounded to three sig figs, the answer will be

`m_(Cu(OH)_2) = color(green)("2.20 g")`