### Solution Found!

# Determining Moles of Released Ions in Dissolution Reactions

**Chapter 4, Problem 13P**

(choose chapter or problem)

**Get Unlimited Answers! Check out our subscriptions**

**QUESTION:**

How many total moles of ions are released when each of the following samples dissolves completely in water?

(a) 0.734 mol of \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\)

(b) 3.86 g of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\)

(c) \(8.66 \times 10^{20}\) formula units of \(\mathrm{NiCl}_{2}\)

#### Watch The Answer!

##### Determining Moles of Released Ions in Dissolution Reactions

Want To Learn More? To watch the entire video and ALL of the videos in the series:

When an ionic compound dissolves in water, it undergoes dissociation into its constituent ions. The total moles of ions released is determined by adding up the moles of each ion generated during this dissociation process. In the case of (a) disodium hydrogen phosphate (Na?HPO?), it dissociates into two sodium ions and one hydrogen phosphate ion. For (b) copper(II) sulfate pentahydrate (CuSO? · 5H?O), it dissociates into one copper ion and one sulfate ion. In (c), nickel(II) chloride (NiCl?) diss

###
Not The Solution You Need? Search for *Your* Answer Here:

### Questions & Answers

**QUESTION:**

How many total moles of ions are released when each of the following samples dissolves completely in water?

(a) 0.734 mol of \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\)

(b) 3.86 g of \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O}\)

(c) \(8.66 \times 10^{20}\) formula units of \(\mathrm{NiCl}_{2}\)

**ANSWER:**

Step 1 of 3

Here we have to calculate how many total moles of ions are released when each of the following samples dissolves completely in water.

(a) \(0.734 \text { mol of } \mathrm{Na}_{2} \mathrm{HPO}_{4}\)

1st we will write a chemical equation for the dissociation of \(\mathrm{K}_{3} \mathrm{PO}_{4}\).

\(\mathrm{Na}_{2} \mathrm{HPO}_{4}(s) \rightarrow 2 \mathrm{Na}^{+}(a q)+\mathrm{HPO}_{4}^{2}(a q)\)

From the above reaction it has been found that 3 moles of ions are released when one mole of \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) dissolves, so the total number of moles released can be calculated as,

\(0.734 \mathrm{~mol} \ \mathrm{Na} \mathrm{HPO}_{4} \times \ 3 \mathrm{~mol} \text { ions } / \mathrm{mol} \mathrm{Na}_{2} \mathrm{HPO}_{4}=1.47 \mathrm{~mol} \text { of ions } \mathrm{Na}^{+} \text {ions }\)

\(0.734 \text { moles of } \mathrm{HPO}_{4}^{2-}\) is also present.

Thus total mole of ion = \(1.47+0.734=2.202 \mathrm{~mol}\)

Thus 2.202 moles of ions are released when \(\mathrm{Na}_{2} \mathrm{HPO}_{4}\) samples dissolves completely in water.