Question

1. write the formulas of the compounds formed by combining the following ions pairs:

a. na⁺ and br⁻ ______
b. fe³⁺ and no₃⁻ ______
c. ba²⁺ and no₂⁻ ______
d. mg²⁺ and po₄³⁻ ______
e. cu²⁺ and co₃²⁻ ______
f. k⁺ and cr₂o₇²⁻ ______
g. fe²⁺ and cn⁻ ______
a. k⁺ and mno₄⁻ ______
b. na⁺ and so₄²⁻ ______
c. nh₄⁺ and c₂o₄²⁻ ______
d. fe³⁺ and po₃³⁻ ______
e. ni²⁺ and br⁻ ______
f. li⁺ and oh⁻ ______
g. nh₄⁺ and s²⁻ ______

Answer:

To solve for the formulas of the compounds formed by combining the given ion pairs, we use the principle of charge neutrality. The total positive charge from the cations must equal the total negative charge from the anions in the compound. This is achieved by finding the least common multiple of the absolute values of the charges of the cation and anion, and then using that to determine the subscripts.

Left Column (First Set of Ion Pairs)

a. $\boldsymbol{\ce{Na+}}$ and $\boldsymbol{\ce{Br-}}$

• Step 1: The charge of $\ce{Na+}$ is $+1$ and the charge of $\ce{Br-}$ is $-1$.
• Step 2: To balance the charges, we need 1 $\ce{Na+}$ and 1 $\ce{Br-}$ because $1\times(+1) + 1\times(-1) = 0$.
• Formula: $\ce{NaBr}$

b. $\boldsymbol{\ce{Fe^{3+}}}$ and $\boldsymbol{\ce{NO3-}}$

• Step 1: The charge of $\ce{Fe^{3+}}$ is $+3$ and the charge of $\ce{NO3-}$ is $-1$.
• Step 2: The least common multiple of 3 and 1 is 3. So we need $3\div3 = 1$ $\ce{Fe^{3+}}$ and $3\div1 = 3$ $\ce{NO3-}$ to balance the charges ($1\times(+3) + 3\times(-1) = 0$).
• Formula: $\ce{Fe(NO3)3}$

c. $\boldsymbol{\ce{Ba^{2+}}}$ and $\boldsymbol{\ce{NO2-}}$

• Step 1: The charge of $\ce{Ba^{2+}}$ is $+2$ and the charge of $\ce{NO2-}$ is $-1$.
• Step 2: The least common multiple of 2 and 1 is 2. So we need $2\div2 = 1$ $\ce{Ba^{2+}}$ and $2\div1 = 2$ $\ce{NO2-}$ to balance the charges ($1\times(+2) + 2\times(-1) = 0$).
• Formula: $\ce{Ba(NO2)2}$

d. $\boldsymbol{\ce{Mg^{2+}}}$ and $\boldsymbol{\ce{PO4^{3-}}}$

• Step 1: The charge of $\ce{Mg^{2+}}$ is $+2$ and the charge of $\ce{PO4^{3-}}$ is $-3$.
• Step 2: The least common multiple of 2 and 3 is 6. So we need $6\div2 = 3$ $\ce{Mg^{2+}}$ and $6\div3 = 2$ $\ce{PO4^{3-}}$ to balance the charges ($3\times(+2) + 2\times(-3) = 0$).
• Formula: $\ce{Mg3(PO4)2}$

e. $\boldsymbol{\ce{Cu^{2+}}}$ and $\boldsymbol{\ce{CO3^{2-}}}$

• Step 1: The charge of $\ce{Cu^{2+}}$ is $+2$ and the charge of $\ce{CO3^{2-}}$ is $-2$.
• Step 2: The least common multiple of 2 and 2 is 2. So we need $2\div2 = 1$ $\ce{Cu^{2+}}$ and $2\div2 = 1$ $\ce{CO3^{2-}}$ to balance the charges ($1\times(+2) + 1\times(-2) = 0$).
• Formula: $\ce{CuCO3}$

f. $\boldsymbol{\ce{K+}}$ and $\boldsymbol{\ce{Cr2O7^{2-}}}$

• Step 1: The charge of $\ce{K+}$ is $+1$ and the charge of $\ce{Cr2O7^{2-}}$ is $-2$.
• Step 2: The least common multiple of 1 and 2 is 2. So we need $2\div1 = 2$ $\ce{K+}$ and $2\div2 = 1$ $\ce{Cr2O7^{2-}}$ to balance the charges ($2\times(+1) + 1\times(-2) = 0$).
• Formula: $\ce{K2Cr2O7}$

g. $\boldsymbol{\ce{Fe^{2+}}}$ and $\boldsymbol{\ce{CN-}}$

• Step 1: The charge of $\ce{Fe^{2+}}$ is $+2$ and the charge of $\ce{CN-}$ is $-1$.
• Step 2: The least common multiple of 2 and 1 is 2. So we need $2\div2 = 1$ $\ce{Fe^{2+}}$ and $2\div1 = 2$ $\ce{CN-}$ to balance the charges ($1\times(+2) + 2\times(-1) = 0$).
• Formula: $\ce{Fe(CN)2}$

Right Column (Second Set of Ion Pairs)

a. $\boldsymbol{\ce{K+}}$ and $\boldsymbol{\ce{MnO4-}}$

• Step 1: The charge of $\ce{K+}$ is $+1$ and the charge of $\ce{MnO4-}$ is $-1$.
• Step 2: To balance the charges, we need 1 $\ce{K+}$ and 1 $\ce{MnO4-}$ because $1\times(+1) + 1\times(-1) = 0$.
• Formula: $\ce{KMnO4}$

b. $\boldsymbol{\ce{Na+}}$ and $\boldsymbol{\ce{SO4^{2-}}}$

• Step 1: The charge of $\ce{Na+}$ is $+1$ and the charge of $\ce{SO4^{2-}}$ is $-2$.
• Step 2: The least common multiple of 1 and 2 is 2. So we need $2\div1 = 2$ $\ce{Na+}$ and $2\div2 = 1$ $\ce{SO4^{2-}}$ to balance the charges ($2\times(+1) + 1\times(-2) = 0$).
• Formula: $\ce{Na2SO4}$

c. $\boldsymbol{\ce{NH4+}}$ and $\boldsymbol{\ce{C2O4^{2-}}}$

• Step 1: The charge of $\ce{NH4+}$ is $+1$ and the charge of $\ce{C2O4^{2-}}$ is $-2$.
• Step 2: The least common multiple of 1 and 2 is 2. So we need $2\div1 = 2$ $\ce{NH4+}$ and $2\div2 = 1$ $\ce{C2O4^{2-}}$ to balance the charges ($2\times(+1) + 1\times(-2) = 0$).
• Formula: $\ce{(NH4)2C2O4}$

d. $\boldsymbol{\ce{Fe^{3+}}}$ and $\boldsymbol{\ce{PO3^{3-}}}$

• Step 1: The charge of $\ce{Fe^{3+}}$ is $+3$ and the charge of $\ce{PO3^{3-}}$ is $-3$.
• Step 2: The least common multiple of 3 and 3 is 3. So we need $3\div3 = 1$ $\ce{Fe^{3+}}$ and $3\div3 = 1$ $\ce{PO3^{3-}}$ to balance the charges ($1\times(+3) + 1\times(-3) = 0$).
• Formula: $\ce{FePO3}$

e. $\boldsymbol{\ce{Ni^{2+}}}$ and $\boldsymbol{\ce{Br-}}$

• Step 1: The charge of $\ce{Ni^{2+}}$ is $+2$ and the charge of $\ce{Br-}$ is $-1$.
• Step 2: The least common multiple of 2 and 1 is 2. So we need $2\div2 = 1$ $\ce{Ni^{2+}}$ and $2\div1 = 2$ $\ce{Br-}$ to balance the charges ($1\times(+2) + 2\times(-1) = 0$).
• Formula: $\ce{NiBr2}$

f. $\boldsymbol{\ce{Li+}}$ and $\boldsymbol{\ce{OH-}}$

• Step 1: The charge of $\ce{Li+}$ is $+1$ and the charge of $\ce{OH-}$ is $-1$.
• Step 2: To balance the charges, we need 1 $\ce{Li+}$ and 1 $\ce{OH-}$ because $1\times(+1) + 1\times(-1) = 0$.
• Formula: $\ce{LiOH}$

g. $\boldsymbol{\ce{NH4+}}$ and $\boldsymbol{\ce{S^{2-}}}$

• Step 1: The charge of $\ce{NH4+}$ is $+1$ and the charge of $\ce{S^{2-}}$ is $-2$.
• Step 2: The least common multiple of 1 and 2 is 2. So we need $2\div1 = 2$ $\ce{NH4+}$ and $2\div2 = 1$ $\ce{S^{2-}}$ to balance the charges ($2\times(+1) + 1\times(-2) = 0$).
• Formula: $\ce{(NH4)2S}$

Final Answers (Left Column)

a. $\ce{NaBr}$
b. $\ce{Fe(NO3)3}$
c. $\ce{Ba(NO2)2}$
d. $\ce{Mg3(PO4)2}$
e. $\ce{CuCO3}$
f. $\ce{K2Cr2O7}$
g. $\ce{Fe(CN)2}$

Final Answers (Right Column)

a. $\ce{KMnO4}$
b. $\ce{Na2SO4}$
c. $\ce{(NH4)2C2O4}$
d. $\ce{FePO3}$
e. $\ce{NiBr2}$
f. $\ce{LiOH}$
g. $\ce{(NH4)2S}$