Chapter 15, Problem 120AP

Interpretation Introduction

(a)

Interpretation:

The number of moles of each ion present in $1.25\text{L}$ of $0.250\text{M}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given as,

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 120AP

The number of moles of ${\text{Na}}^{+}$ and ${\text{PO}}_4^{3-}$ ions present in $1.25\text{L}$ of $0.250\text{M}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $0.9375\text{mol}$ and $0.3125\text{mol}$.

Explanation of Solution

The molarity of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $0.250\text{M}$.

The volume of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $1.25\text{L}$.

The number of moles of a solute present in the solution is given as,

$\text{n}=\text{M}\text{V}$

Where,

• $\text{M}$ represents molarity of the solution.
• $\text{V}$ represents the volume of the solution.

Substitute the value of molarity and volume of the ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution in the above equation.

$\begin{array}{c}\text{n}=\left(0.250\text{M}\right)\left(1.25\text{L}\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1\text{M}}\right)\\ =0.3125\text{mol}\end{array}$

The number of moles of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ present in the given solution is $0.3125\text{mol}$.

The dissociation of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ in water is represented as,

${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}\left(\text{a}\text{q}\right)\to 3{\text{Na}}^{+}\left(\text{a}\text{q}\right)+{\text{PO}}_4^{3-}\left(\text{a}\text{q}\right)$

One mole of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ produces three moles of ${\text{Na}}^{+}$ ions. Hence, $0.3125\text{mol}$ of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ will produce $0.9375\text{mol}$ of ${\text{Na}}^{+}$ ions.

One mole of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ produces one mole of ${\text{PO}}_4^{3-}$ ions. Hence, $0.3125\text{mol}$ of ${\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ will produce $0.3125\text{mol}$ of ${\text{PO}}_4^{3-}$ ions.

Therefore, the number of moles of ${\text{Na}}^{+}$ and ${\text{PO}}_4^{3-}$ ions present in $1.25\text{L}$ of $0.250\text{M}{\text{Na}}_{\text{3}}{\text{PO}}_{\text{4}}$ solution is $0.9375\text{mol}$ and $0.3125\text{mol}$.respectively.

Interpretation Introduction

(b)

Interpretation:

The number of moles of each ion present in $3.5\text{mL}$ of $6.0\text{M}{\text{H}}_2{\text{SO}}_{\text{4}}$ solution is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given as,

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 120AP

The number of moles of ${\text{H}}^{+}$ and ${\text{SO}}_4^{2-}$ ions present in $3.5\text{mL}$ of $6.0\text{M}{\text{H}}_2{\text{SO}}_{\text{4}}$ solution is $0.042\text{mol}$ and $0.021\text{mol}$ respectively.

Explanation of Solution

The molarity of ${\text{H}}_2{\text{SO}}_{\text{4}}$ solution is $6.0\text{M}$.

The volume of ${\text{H}}_2{\text{SO}}_{\text{4}}$ solution is $3.5\text{mL}$.

The number of moles of a solute present in the solution is given as,

$\text{n}=\text{M}\text{V}$

Where,

• $\text{M}$ represents molarity of the solution.
• $\text{V}$ represents the volume of the solution.

Substitute the value of molarity and volume of the ${\text{H}}_2{\text{SO}}_{\text{4}}$ solution in the above equation.

$\begin{array}{c}\text{n}=\left(6.0\text{M}\right)\left(3.5\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1\text{M}}\right)\\ =0.021\text{mol}\end{array}$

The number of moles of ${\text{H}}_2{\text{SO}}_{\text{4}}$ present in the given solution is $0.021\text{mol}$.

The dissociation of ${\text{H}}_2{\text{SO}}_{\text{4}}$ in water is represented as,

${\text{H}}_2{\text{SO}}_{\text{4}}\left(\text{a}\text{q}\right)\to 2{\text{H}}^{+}\left(\text{a}\text{q}\right)+{\text{SO}}_4^{2-}\left(\text{a}\text{q}\right)$

One mole of ${\text{H}}_2{\text{SO}}_{\text{4}}$ produces two moles of ${\text{H}}^{+}$ ion. Hence, $0.021\text{mol}$ of ${\text{H}}_2{\text{SO}}_{\text{4}}$ will dissociate in $0.042\text{mol}$ of ${\text{H}}^{+}$ ions.

One mole of ${\text{H}}_2{\text{SO}}_{\text{4}}$ produces one mole of ${\text{SO}}_4^{2-}$ ion. Hence, $0.021\text{mol}$ of ${\text{H}}_2{\text{SO}}_{\text{4}}$ will dissociate in $0.021\text{mol}$ of ${\text{SO}}_4^{2-}$ ions.

Therefore, the number of moles of ${\text{H}}^{+}$ and ${\text{SO}}_4^{2-}$ ions present in $3.5\text{mL}$ of $6.0\text{M}{\text{H}}_2{\text{SO}}_{\text{4}}$ solution is $0.042\text{mol}$ and $0.021\text{mol}$ respectively.

Interpretation Introduction

(c)

Interpretation:

The number of moles of each ion present in $25\text{mL}$ of $0.15\text{M}{\text{AlCl}}_3$ solution is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given as,

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 120AP

The number of moles of ${\text{Al}}^{3+}$ and ${\text{Cl}}^{-}$ ion present in $25\text{mL}$ of $0.15\text{M}{\text{AlCl}}_3$ solution is is $3.75\times {10}^{-3}\text{mol}$ and $1.125\times {10}^{-2}\text{mol}$ respectively.

Explanation of Solution

The molarity of ${\text{AlCl}}_3$ solution is $0.15\text{M}$.

The volume of ${\text{AlCl}}_3$ solution is $25\text{mL}$.

The number of moles of a solute present in the solution is given as,

$\text{n}=\text{M}\text{V}$

Where,

• $\text{M}$ represents molarity of the solution.
• $\text{V}$ represents the volume of the solution.

Substitute the value of molarity and volume of the ${\text{AlCl}}_3$ solution in the above equation.

$\begin{array}{c}\text{n}=\left(0.15\text{M}\right)\left(25\text{mL}\right)\left(\frac{1\text{L}}{1000\text{mL}}\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1\text{M}}\right)\\ =3.75\times {10}^{-3}\text{mol}\end{array}$

The number of moles of ${\text{AlCl}}_3$ present in the given solution is $3.75\times {10}^{-3}\text{mol}$.

The dissociation of ${\text{AlCl}}_3$ in water is represented as,

${\text{AlCl}}_3\left(\text{a}\text{q}\right)\to {\text{Al}}^{3+}\left(\text{a}\text{q}\right)+3{\text{Cl}}^{-}\left(\text{a}\text{q}\right)$

One mole of ${\text{AlCl}}_3$ produces one mole of ${\text{Al}}^{3+}$ ion. Hence, $3.75\times {10}^{-3}\text{mol}$ of ${\text{AlCl}}_3$ will dissociate in $3.75\times {10}^{-3}\text{mol}$ of ${\text{Al}}^{3+}$ ions.

One mole of ${\text{AlCl}}_3$ produces three moles of ${\text{Cl}}^{-}$ ion. Hence, $3.75\times {10}^{-3}\text{mol}$ of ${\text{AlCl}}_3$ will dissociate in $1.125\times {10}^{-2}\text{mol}$ of ${\text{Cl}}^{-}$ ions.

Therefore, the number of moles of ${\text{Al}}^{3+}$ and ${\text{Cl}}^{-}$ ion present in $25\text{mL}$ of $0.15\text{M}{\text{AlCl}}_3$ solution is is $3.75\times {10}^{-3}\text{mol}$ and $1.125\times {10}^{-2}\text{mol}$ respectively.

Interpretation Introduction

(d)

Interpretation:

The number of moles of each ion present in $1.50\text{L}$ of $1.25\text{M}{\text{BaCl}}_{\text{2}}$ solution is to be calculated.

Concept Introduction:

The molarity of a solution is defined as the number of mole of solute dissolved in one liter of the solution. The formula for molarity is given as,

$\text{M}=\frac{\text{n}}{\text{V}}$

Where,

• $\text{n}$ represents the number of moles of the solute.
• $\text{V}$ represents the volume of the solution.

Answer to Problem 120AP

The number of moles of ${\text{Ba}}^{2+}$ and ${\text{Cl}}^{-}$ ion present in $1.50\text{L}$ of $1.25\text{M}{\text{BaCl}}_{\text{2}}$ solution is $1.875\text{mol}$ and $3.75\text{mol}$ respectively.

Explanation of Solution

The molarity of ${\text{BaCl}}_{\text{2}}$ solution is $1.25\text{M}$.

The volume of ${\text{BaCl}}_{\text{2}}$ solution is $1.50\text{L}$.

The number of moles of a solute present in the solution is given as,

$\text{n}=\text{M}\text{V}$

Where,

• $\text{M}$ represents molarity of the solution.
• $\text{V}$ represents the volume of the solution.

Substitute the value of molarity and volume of the ${\text{BaCl}}_{\text{2}}$ solution in the above equation.

$\begin{array}{c}\text{n}=\left(1.25\text{M}\right)\left(1.50\text{L}\right)\left(\frac{1\text{mol}\cdot {\text{L}}^{-1}}{1\text{M}}\right)\\ =1.875\text{mol}\end{array}$

The number of moles of ${\text{BaCl}}_{\text{2}}$ present in the given solution is $1.875\text{mol}$.

The dissociation of ${\text{BaCl}}_{\text{2}}$ in water is represented as,

${\text{BaCl}}_{\text{2}}\left(\text{a}\text{q}\right)\to {\text{Ba}}^{2+}\left(\text{a}\text{q}\right)+2{\text{Cl}}^{-}\left(\text{a}\text{q}\right)$

One mole of ${\text{BaCl}}_{\text{2}}$ produces one mole of ${\text{Ba}}^{2+}$ ions. Hence, $1.875\text{mol}$ of ${\text{BaCl}}_{\text{2}}$ will dissociate in $1.875\text{mol}$ of ${\text{Ba}}^{2+}$ ions.

One mole of ${\text{BaCl}}_{\text{2}}$ produces two moles of ${\text{Cl}}^{-}$ ions. Hence, $1.875\text{mol}$ of ${\text{BaCl}}_{\text{2}}$ will dissociate in $3.75\text{mol}$ of ${\text{Cl}}^{-}$ ions.

Therefore, the number of moles of ${\text{Ba}}^{2+}$ and ${\text{Cl}}^{-}$ ion present in $1.50\text{L}$ of $1.25\text{M}{\text{BaCl}}_{\text{2}}$ solution is $1.875\text{mol}$ and $3.75\text{mol}$ respectively.