# 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?

Oct 3, 2015

$\text{2.20 g}$

#### 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

$C = \frac{n}{V} \implies n = C \cdot V$

${n}_{O {H}^{-}} = {\text{0.450 M" * 100 * 10^(-3)"L" = "0.0450 moles OH}}^{-}$

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, $\textcolor{red}{2}$ 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

${m}_{C u {\left(O H\right)}_{2}} = \textcolor{g r e e n}{\text{2.20 g}}$