Question

Calculate the $\left[{\mathrm{OH}}^{-}\right]$ in each of the following solutions, and indicate whether the solution is acidic or basic. a. $\left[{\mathrm{H}}^{+}\right]=1.02\times {10}^{-7}M$ b. $\left[{\mathrm{H}}^{+}\right]=9.77\times {10}^{-8}M$ c. $\left[{\mathrm{H}}^{+}\right]=3.41\times {10}^{-3}M$ d. $\left[{\mathrm{H}}^{+}\right]=4.79\times {10}^{-11}M$

Short answer

a) $[{\mathrm{OH}}^{-}]=9.8\times {10}^{-8}\mathrm{M}$, acidic b) $[{\mathrm{OH}}^{-}]=1.02\times {10}^{-7}\mathrm{M}$, basic c) $[{\mathrm{OH}}^{-}]=2.93\times {10}^{-12}\mathrm{M}$, acidic d) $[{\mathrm{OH}}^{-}]=2.08\times {10}^{-4}\mathrm{M}$, basic

Step-by-step solution

Find the ion product of water (Kw)

The ion product of water ($K_{\mathrm{w}}$) is a constant value equal to $1\times {10}^{-14}$ at 25°C.

Calculate $[{\mathrm{OH}}^{-}]$ using the given $[{\mathrm{H}}^{+}]$

To find the hydroxide ion concentration $[{\mathrm{OH}}^{-}]$, use the equation: $K_{\mathrm{w}}=[{\mathrm{H}}^{+}][{\mathrm{OH}}^{-}]$. For each solution, solve for $[{\mathrm{OH}}^{-}]$ and determine if it's acidic or basic: a. $1.02\times {10}^{-7}M$ of $[{\mathrm{H}}^{+}]$:

a) Solve for $[{\mathrm{OH}}^{-}]$ in solution a

$[{\mathrm{OH}}^{-}]=\frac{K_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}=\frac{1\times {10}^{-14}}{1.02\times {10}^{-7}M}=9.8\times {10}^{-8}\mathrm{M}$ b. $9.77\times {10}^{-8}M$ of $[{\mathrm{H}}^{+}]$:

b) Solve for $[{\mathrm{OH}}^{-}]$ in solution b

$[{\mathrm{OH}}^{-}]=\frac{K_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}=\frac{1\times {10}^{-14}}{9.77\times {10}^{-8}M}=1.02\times {10}^{-7}\mathrm{M}$ c. $3.41\times {10}^{-3}M$ of $[{\mathrm{H}}^{+}]$:

c) Solve for $[{\mathrm{OH}}^{-}]$ in solution c

$[{\mathrm{OH}}^{-}]=\frac{K_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}=\frac{1\times {10}^{-14}}{3.41\times {10}^{-3}M}=2.93\times {10}^{-12}\mathrm{M}$ d. $4.79\times {10}^{-11}M$ of $[{\mathrm{H}}^{+}]$:

d) Solve for $[{\mathrm{OH}}^{-}]$ in solution d

$[{\mathrm{OH}}^{-}]=\frac{K_{\mathrm{w}}}{[{\mathrm{H}}^{+}]}=\frac{1\times {10}^{-14}}{4.79\times {10}^{-11}M}=2.08\times {10}^{-4}\mathrm{M}$ Step 2: Determine if the solution is acidic or basic

Compare $[{\mathrm{H}}^{+}]$ and $[{\mathrm{OH}}^{-}]$ to find the nature of the solution

a) $[{\mathrm{H}}^{+}]=1.02\times {10}^{-7}\mathrm{M}$ and $[{\mathrm{OH}}^{-}]=9.8\times {10}^{-8}\mathrm{M}$: $[{\mathrm{H}}^{+}]>[{\mathrm{OH}}^{-}]$, the solution is acidic. b) $[{\mathrm{H}}^{+}]=9.77\times {10}^{-8}\mathrm{M}$ and $[{\mathrm{OH}}^{-}]=1.02\times {10}^{-7}\mathrm{M}$: $[{\mathrm{H}}^{+}]<[{\mathrm{OH}}^{-}]$, the solution is basic. c) $[{\mathrm{H}}^{+}]=3.41\times {10}^{-3}\mathrm{M}$ and $[{\mathrm{OH}}^{-}]=2.93\times {10}^{-12}\mathrm{M}$: $[{\mathrm{H}}^{+}]>[{\mathrm{OH}}^{-}]$, the solution is acidic. d) $[{\mathrm{H}}^{+}]=4.79\times {10}^{-11}\mathrm{M}$ and $[{\mathrm{OH}}^{-}]=2.08\times {10}^{-4}\mathrm{M}$: $[{\mathrm{H}}^{+}]<[{\mathrm{OH}}^{-}]$, the solution is basic.

Key concepts

Acidic and Basic Solutions

Understanding the nature of solutions as acidic or basic is fundamental in chemistry. A solution is considered acidic when the concentration of hydrogen ions $[{\mathrm{H}}^{+}]$ is greater than that of hydroxide ions $[{\mathrm{OH}}^{-}]$. Conversely, a solution is basic when $[{\mathrm{OH}}^{-}]$ exceeds $[{\mathrm{H}}^{+}]$. Neutral solutions occur when these concentrations are equal. The degree of acidity or basicity of a solution can be measured using a pH scale, which is a topic we'll cover shortly.

For example, in our exercise, by comparing the given $[{\mathrm{H}}^{+}]$ and the calculated $[{\mathrm{OH}}^{-}]$ concentrations, we determine whether the solution is acidic or basic. This is critical as the properties of a solution – including reactivity, taste, and biological availability – vary significantly with pH.

Ion Product of Water

The ion product of water ($K_w$) is the product of the molar concentrations of hydrogen and hydroxide ions in water at a specific temperature, typically 25°C. The constant value $K_w=1\times {10}^{-14}$ has profound implications. It signifies a state of dynamic equilibrium in pure water, where water molecules dissociate into ions and simultaneously recombine.

In the context of our exercise, $K_w$ allows us to calculate the unknown hydroxide ion concentration. By using the equation $K_w=[{\mathrm{H}}^{+}][{\mathrm{OH}}^{-}]$, and given the concentration of hydrogen ions, the concentration of hydroxide ions can be solved for each scenario. Taking this further, we can also infer the pH and pOH of the solution, leading us directly to our next concept.

pH and pOH Calculations

The pH is a logarithmic scale used to specify the acidity or basicity of an aqueous solution, and it is defined as the negative base 10 logarithm of the activity of hydrogen ions. The equation $\text{pH}=-\log ([{\mathrm{H}}^{+}])$ allows us to convert the hydrogen ion concentration into a value that's easier to work with.

The pOH works similarly for hydroxide ions and is defined as $\text{pOH}=-\log ([{\mathrm{OH}}^{-}])$. Knowing either the pH or pOH of a solution allows us to determine the other since $\text{pH}+\text{pOH}=14$, at 25°C. This interrelationship is utilized to calculate unknowns in different scenarios and can be highlighted as a practical application of logarithms in chemistry.

Chemical Equilibrium

Chemical equilibrium occurs when the rates of the forward and reverse reactions in a chemical system are equal, resulting in no net change in the concentrations of the reactants and products over time. This concept is essential when considering reactions involving the dissociation of water or weak acids and bases.

In our exercise, we assume equilibrium in the ion product of water to calculate $[{\mathrm{OH}}^{-}]$. The concept of equilibrium is also central to understanding buffer systems, solubility, and reaction kinetics. Therefore, recognizing when a system is at equilibrium and how it can be affected (for example, by changes in concentration, temperature, or pressure) is fundamental in predicting the behavior of chemical reactions.