# Common Ion Effect

- Page ID
- 36298

Skills to Develop

- Recognize common ions from various salts, acids, and bases.
- Calculate concentrations involving common ions.
- Calculate ion concentrations involving chemical equilibrium.
- Apply the Newton's method to solve cubic equations.

The solubility products *K*_{sp}'s are equilibrium constants in hetergeneous equilibria. If several salts are present in a system, they all ionize in the solution. If the salts contain a common cation or anion, these salts contribute to the concentration of the common ion. Contributions from all salts must be included in the calculation of concentration of the common ion. For example, a solution containing sodium chloride and potassium chloride will have the following relationship:

\(\mathrm{[Na^+] + [K^+] = [Cl^-]}\)

Consideration of *charge balance* or *mass balance* or both leads to the same conclusion.

## Common Ions

When \(\ce{NaCl}\) and \(\ce{KCl}\) are dissolved in the same solution, the \(\mathrm{ {\color{Green} Cl^-}}\) ions are *common *to both salts. In a system containing \(\ce{NaCl}\) and \(\ce{KCl}\), the \(\mathrm{ {\color{Green} Cl^-}}\) ions are common ions.

\[\mathrm{NaCl \rightleftharpoons Na^+ + {\color{Green} Cl^-}}\]

\[\mathrm{KCl \rightleftharpoons K^+ + {\color{Green} Cl^-}}\]

\[\mathrm{CaCl_2 \rightleftharpoons Ca^{2+} + {\color{Green} 2 Cl^-}}\]

\[\mathrm{AlCl_3 \rightleftharpoons Al^{3+} + {\color{Green} 3 Cl^-}}\]

\[\mathrm{AgCl \rightleftharpoons Ag^+ + {\color{Green} Cl^-}}\]

For example, when \(\ce{AgCl}\) is dissolved into a solution already containing \(\ce{NaCl}\) (actually \(\ce{Na+}\) and \(\ce{Cl-}\) ions), the \(\ce{Cl-}\) ions come from the ionization of both \(\ce{AgCl}\) and \(\ce{NaCl}\). Thus, \(\ce{[Cl- ]}\) differs from \(\ce{[Ag+]}\). The following examples show how the concentration of the common ion is calculated.

Example \(\PageIndex{1}\)

What are \(\ce{[Na+]}\), \(\ce{[Cl- ]}\), \(\ce{[Ca^2+]}\), and \(\ce{[H+]}\) in a solution containing 0.10 M each of \(\ce{NaCl}\), \(\ce{CaCl2}\), and \(\ce{HCl}\)?

**SOLUTION**

Due to the conservation of ions, we have

\(\mathrm{[Na^+] = [Ca^{2+}] = [H^+] = 0.10\: \ce M}\).

but

\(\begin{alignat}{3}

&\ce{[Cl- ]} &&= && && \:\textrm{0.10 (due to NaCl)}\\

& && && + &&\mathrm{\:0.20\: (due\: to\: CaCl_2)}\\

& && && + &&\mathrm{\:0.10\: (due\: to\: HCl)}\\

& &&= && &&\mathrm{\:0.40\: M}

\end{alignat}\)

Exercise \(\PageIndex{1}\)

John poured 10.0 mL of 0.10 M \(\ce{NaCl}\), 10.0 mL of 0.10 M \(\ce{KOH}\), and 5.0 mL of 0.20 M \(\ce{HCl}\) solutions together and then he made the total volume to be 100.0 mL. What is \(\ce{[Cl- ]}\) in the final solution?

\(\mathrm{[Cl^-] = \dfrac{0.1\: M\times 10\: mL+0.2\: M\times 5.0\: mL}{100.0\: mL} = 0.020\: M}\)

Example \(\PageIndex{2}\)

What is the solubility (in moles per liter) of \(\ce{AgCl}\) in a solution that is 0.05 M in \(\ce{KCl}\)? *K*_{sp} for \(\ce{AgCl}\) is 1.0E-10.

**SOLUTION**

Assume \(x = \ce{[Ag+]}\) to be the molar solubility, then we can write the equilibrium concentration below the formula of the equilibrium.

\(\begin{array}{cccccl}

\ce{AgCl &\rightarrow &Ag+ &+ &Cl- &}\\

&&x &&0.05+x &\leftarrow \ce{equilibrium\:\, concentration}

\end{array}\)

By definition of the *K*_{sp}, we have

\(x (0.05 + x) = K_{\ce{sp}}\);

Thus,

\(x = \dfrac{\textrm{1.0e-10}}{0.05} = \textrm{2e-9 M}\)

Note that 0.05 + x is approximately 0.05, and you may use this approximation for the calculation.

Example \(\PageIndex{3}\)

What is the solubility of \(\ce{Li2CO3}\) in a solution also containing 0.10 M of \(\ce{K2CO3}\)? The solubility product for \(\ce{Li2CO3}\) is 0.0017.

**SOLUTION**

In this case, \(\ce{CO3^2-}\) is the common ion between the two salts.

\(\ce{K2CO3 \rightleftharpoons 2 K+ + CO3^2-}\)

\(\ce{Li2CO3 \rightleftharpoons 2 Li+ + CO3^2-}\)

This question implies that \(\ce{K2CO3}\) is soluble. Thus, the only heterogeneous equilibrium to be considered is

\(\ce{K2CO3 \rightleftharpoons 2 K+ + CO3^2-}\)

Let \(\ce{[Li+]} = 2 x\), then \(\ce{[CO3^2- ]} = 0.10 + x\). Thus, we can write down the equilibrium equation again, and write the concentrations below the chemical formula:

\[\begin{array}{ccccc}

\ce{Li2CO3 &\rightleftharpoons &2 Li+ &+ &CO3^2-}\\

&&2 x &&0.10+x

\end{array}\]

\[\ce{[Li+]^2 [CO3^2- ]} = (2 x)^2 (0.10+x) = 0.0017\]

Expanding the formula results in

\[x^3 + 0.10 x^2 - 0.00043 = 0\.]

A general method to solve this equation is to use the Newton's method. For which, we assume that

\[y = x^3 + 0.10 x^2 - 0.00043 = 0.\]

We assume the value of \(x = 0.06\), and substitute in the equation to calculate y.

\[y = 0.06^3 + 0.10\times 0.06^2 - 0.00043 = 0.000146\]

We now assume \(x = 0.05\), and substitute in the equation to evaluate y,

\[y = 0.05^3 + 0.10\times 0.05^2 - 0.00043 = -0.000055\]

Since the values for y calculated for x = 0.05 and 0.06 have different sign, the x value should lie between 0.05 and 0.06. We further assume the value for x = 0.053, and we obtain a y value

\[y = 0.053^3 + 0.10\times 0.053^2 - 0.00043 = -2.2\times 10^{-7}\]

Thus, the value of x should be greater than 0.053. We can increase x to 0.0531 or 0.0532, and evaluate y again:

\[y = 0.0531^3 + 0.10\times 0.0531^2 - 0.00043 = 1.7\times 10^{-6}\]

Thus, x value should lie between 0.0531 and 0.0530. By this method, we have evaluated x values to three significant figures. We only need two significant figures due to the nature of the data in the problem. Thus we have

\[\ce{[Li+]} = 2 x = \textrm{0.106 \;M}\]

**DISCUSSION**

The molar solubility of \(\ce{Li2CO3}\) in a solution also containing 0.10 M of \(\ce{K2CO3}\) is 0.053 mole per liter, but

\[\ce{[Li+]} = 2 x = \textrm{0.106 M}\]

This example illustrates the Newton's method for solving cubic equations. Note that if the *K*_{sp} is small, then x is a very small value. In this case, \(0.10 + x \approx 0.10\). You do not need to use the Newton's method in this case.

Example \(\PageIndex{4}\)

What is the solubility of \(\ce{Li2CO3}\) in a solution which is 0.20 M in \(\ce{Na2CO3}\)? The solubility product of \(\ce{Li2CO3}\) is 1.7e-3 M^{3}.

**SOLUTION**

Since the solution contains 0.20 M \(\ce{Na2CO3}\), \(\ce{[CO3^2- ]} = \textrm{0.20 M}\). Assume the solubility to be *x* M of \(\ce{Li2CO3}\), then we have

\(\begin{array}{ccccc}

\ce{Li2CO3 &\rightleftharpoons &2 Li+ &+ &CO3^2-}\\

&&2 x &&x + 0.20\:\ce M

\end{array}\)

Thus,

\(\begin{align}

(2 x)^2 (x+0.20) &= \textrm{1.7e-3 M}^3\\

4x^3 + 0.20 x^2 &= \textrm{1.7e-3 M}^3

\end{align}\)

\(x^3 + 0.20 x^2 - 0.00043 = 0\)

There is no definite way to solve this equation, but the Newton's method is very useful. In this method, we let

\(y = x^3 + 0.20 x^2 - 0.00043\)

We guesstimate *x* and evaluate *y* using the above expression. The solution is a value for *x* so that *y*=0. We refine *x* values between two *x* values that give positive and negative *y* values in progression as shown in the Table below.

y |
x |
Remarks |
---|---|---|

\(2 \times 10^{4}\) | 0.05 | any small value for x |

\(5\times 10^{-4}\) | 0.06 | y larger, try x < 0.05 |

\(-5\times 10^{-5}\) | 0.04 | 0.04 < x < 0.05 |

\(8\times 10^{-6}\) | 0.0425 | 0.04 < x < 0.0425 |

\(-1\times 10^{-5}\) | 0.0415 | 0.0415 < x < 0.0425 |

\(3\times 10^{-6}\) | 0.0420 | 0.0415 < x < 0.0420 |

\(-5\times 10^{-6}\) | 0.0419 | 0.0415 < x < 0.0419 |

**DISCUSSION**

This is the same type of question as **Example 3**. As the carbonate concentration increases from 0.10 to 0.20 M, the solubility of lithium carbonate reduces to 0.042 M from 0.053 M. Note that the lithium ion concentration is 0.084 M in this case.

Note that when used to treat depression, lithium carbonate is usually called lithium in the health care profession.