Chapter 2, Problem 61QP

Name the following compounds: $\begin{array}{l}\left(\text{a}\right){\text{ KH}}_{\text{2}}{\text{PO}}_{\text{4}}\text{, }\left(\text{b}\right){\text{ K}}_{\text{2}}{\text{HPO}}_{\text{4}}\text{, }\left(\text{c}\right)\text{ HBr }\left(\text{gas}\right)\text{, }\left(\text{d}\right)\text{ HBr }\left(\text{in water}\right)\text{, }\left(\text{e}\right){\text{ Li}}_{\text{2}}{\text{CO}}_{\text{3}}\text{,}\\ \left(\text{f}\right){\text{ K}}_{\text{2}}{\text{Cr}}_{\text{2}}{\text{O}}_{\text{7}}\text{, }\left(\text{g}\right){\text{ NH}}_{\text{4}}{\text{NO}}_{\text{2}}\text{, }\left(\text{h}\right){\text{ HIO}}_{\text{3}}\text{, }\left(\text{i}\right){\text{ PF}}_{\text{3}}\text{, }\left(\text{j}\right){\text{ P}}_{\text{4}}{\text{O}}_{\text{6}}\text{, }\left(\text{k}\right){\text{ Cdl}}_{\text{2}}\text{,}\left(\text{1}\right){\text{ SrSO}}_{\text{4}}\text{, }\left(\text{m}\right)\text{ Al}{\left(\text{OH}\right)}_{\text{3}}\text{.}\end{array}$

Interpretation Introduction

Interpretation:

The names the of given compounds are to be identified from the formula.

Concept introduction:

Steps for naming the ionic compounds are as follows:

Identify the cation and the anion in the compound.

Combine the names of the cations and the anions by elimination of the word ‘ion’ from the names of the individual ions.

Steps for naming the binary molecular compounds are as follows:

First name the element that appears first in its molecular formula.

Now, name the second element by changing its ending to –ide.

Use proper prefixes to indicate the number of atoms present for individual elements.

Steps for naming the acids are as follows:

From hydrogen, omit the term ‘–gen’, leaving the name hydro-.

The name ending with ‘–ide’ is changed to ‘–ic’.

Now, combine the two terms and add the word acid to give complete name.

The oxoacids are named according to the names of the oxoanions from which they are derived. The acid based on an -ate ion is called –ic acid.

Answer to Problem 61QP

Solution:

a)

Potassium dihydrogen phosphate

b)

Potassium hydrogen phosphate

c)

Hydrogen bromide

d)

Hydrobromic acid

e)

Lithium carbonate

f)

Potassium dichromate

g)

Ammonium nitrite

h)

Hydrogen iodate (in water Iodic acid)

i)

Phosphorous pentafluoride

j)

Tetraphosphorus hexoxide

k)

Cadmium iodide

l)

Strontium sulfate

m)

Aluminum hydroxide

Explanation of Solution

a) ${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$

The compound ${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$ contains a potassium ion $\left({\text{K}}^{\text{+}}\right)$ and a dihydrogen phosphate ion $\left({\text{H}}_{\text{2}}{\text{PO}}_4{}^{-}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{KH}}_{\text{2}}{\text{PO}}_{\text{4}}$ is potassium dihydrogen phosphate.

b) ${\text{K}}_2{\text{HPO}}_{\text{4}}$

The compound ${\text{K}}_2{\text{HPO}}_{\text{4}}$ a contains potassium ion $\left({\text{K}}^{\text{+}}\right)$ and a hydrogen phosphate ion $\left({\text{HPO}}_4{}^{2-}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{K}}_2{\text{HPO}}_{\text{4}}$ is potassium hydrogen phosphate.

c) $\text{HBr}$

The compound $\text{HBr}$ contains a hydrogen ion $\left({\text{H}}^{\text{+}}\right)$ and a bromide ion $\left({\text{Br}}^{\text{-}}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of $\text{HBr}$ is hydrogen bromide.

d) $\text{HBr}\left(\text{in}\text{water}\right)$

The molecular compound $\text{HBr}$ is a simple acid. From the word hydrogen, omit ‘–gen’ from it leaving only hydro-. Also, change the ‘–ide’ of the term bromide to ‘–ic’ forming the name ‘–bromic’. Combine the two words and add the word ‘acid’. The name of $\text{HBr}\left(\text{in}\text{water}\right)$ is hydrobromic acid.

e) ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$

The compound ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ contains a lithium ion $\left({\text{Li}}^{\text{+}}\right)$ and a carbonate ion $\left({\text{CO}}_3{}^{2-}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{Li}}_{\text{2}}{\text{CO}}_{\text{3}}$ is lithium carbonate.

f) ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$

The compound ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ contains a potassium ion $\left({\text{K}}^{\text{+}}\right)$ and a dichromate ion $\left({\text{Cr}}_2{\text{O}}_7{}^{2-}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ is potassium dichromate.

g) ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{2}}$

The compound ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{2}}$ contains an ammonium ion $\left({\text{NH}}_4{}^{\text{+}}\right)$ and a nitrite ion $\left({\text{NO}}_{\text{2}}{}^{\text{-}}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{2}}$ is ammonium nitrite.

h) ${\text{HIO}}_3$

The compound ${\text{HIO}}_3$ is an oxoacid, which is based on iodate ion $\left({\text{IO}}_3{}^{-}\right)$ and has one ionizable hydrogen atom $\left({\text{H}}^{\text{+}}\right)$. The name of ${\text{HIO}}_3$ in water is iodic acid. The actual name of the compound is hydrogen iodate.

i) ${\text{PF}}_{\text{5}}$

The molecular compound ${\text{PF}}_{\text{5}}$ contains one phosphorous atom and five fluorine atoms. The element that appears first in the formula is phosphorous. Omit the prefix mono- for phosphorous. The second element is fluorine, so change fluorine to fluoride and use the prefix penta- to denote the number of fluorine atoms. The name of ${\text{PF}}_{\text{5}}$ is Phosphorous pentafluoride.

j) ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$

The molecular compound ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ contains four phosphorus atoms and six oxygen atoms. The element that appears first in the formula is phosphorous. Use the prefix tetra- to denote the number of phosphorous atoms. The second element is oxygen, so change oxygen to oxide and use the prefix hexa- to denote the number of oxygen atoms. The name of ${\text{P}}_{\text{4}}{\text{O}}_{\text{6}}$ is Tetraphosphorus hexoxide.

k) ${\text{CdI}}_{\text{2}}$

The compound ${\text{CdI}}_{\text{2}}$ contains cadmium ion $\left({\text{Cd}}^2{}^{\text{+}}\right)$ and iodide ion $\left({\text{I}}^{\text{-}}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{CdI}}_{\text{2}}$ is cadmium iodide.

l) ${\text{SrSO}}_4$

The compound ${\text{SrSO}}_4$ contains a strontium ion $\left({\text{Sr}}^{\text{2}}{}^{\text{+}}\right)$ and a sulfate ion $\left({\text{SO}}_{\text{4}}{}^{\text{2-}}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of ${\text{SrSO}}_4$ is strontium sulfate.

m) $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$

The compound $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ contains an aluminum ion $\left({\text{Al}}^{\text{3}}{}^{\text{+}}\right)$ and a hydroxide ion $\left({\text{OH}}^{\text{-}}\right)$. Combine the cation and the anion names, and eliminate the word ion from each of the individual ion names, the name of $\text{Al}{\left(\text{OH}\right)}_{\text{3}}$ is aluminum hydroxide.