Unlocking Weak Base Equilibria: How to Find Hydroxide Concentration in Ammonia

Understanding how weak bases like ammonia (NH₃) behave in water is fundamental in chemistry. Unlike strong bases that dissociate completely, weak bases only partially ionize, establishing an equilibrium. This guide will walk you through calculating the hydroxide ion concentration ([OH⁻]) in a 0.0750 M ammonia solution, focusing specifically on the detailed computation to arrive at the intermediate value \(x^2 = 1.31 \times 10^{-6}\), as requested.

Key Insights: Calculating [OH⁻] for Ammonia

Equilibrium Matters: Ammonia is a weak base, meaning its reaction with water \((\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-)\) reaches an equilibrium described by the base ionization constant, \(K_b\).
The \(K_b\) Expression: The equilibrium concentrations are related by \(K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}\). For NH₃, \(K_b = 1.75 \times 10^{-5}\).
Approximation Simplifies Calculation: Because \(K_b\) is small, we can often approximate \([\text{NH}_3]_{equilibrium} \approx [\text{NH}_3]_{initial}\), leading to \(K_b \approx \frac{x^2}{[\text{NH}_3]_{initial}}\), where \(x = [\text{OH}^-]\).

The Chemistry: Ammonia as a Weak Base

Understanding the Ionization Reaction

When ammonia dissolves in water, it acts as a Brønsted-Lowry base by accepting a proton (H⁺) from a water molecule. This process forms the ammonium ion (NH₄⁺) and the hydroxide ion (OH⁻). The reaction establishes an equilibrium, as shown below:

\[ \text{NH}_3 (aq) + \text{H}_2\text{O} (l) \rightleftharpoons \text{NH}_4^+ (aq) + \text{OH}^- (aq) \]

The extent to which this reaction proceeds to the right is quantified by the base ionization constant, \(K_b\). A small \(K_b\) value indicates that only a small fraction of the base molecules ionize.

Chemical structure and reaction of Ammonium Hydroxide

Depiction of the equilibrium involving ammonia, water, ammonium ions, and hydroxide ions.

The Base Ionization Constant (\(K_b\))

The equilibrium expression for the ionization of ammonia is:

\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} \]

You are given:

Initial concentration of NH₃, \([\text{NH}_3]_{initial} = 0.0750 \, \text{M}\)
Base ionization constant for NH₃, \(K_b = 1.75 \times 10^{-5}\)

Our goal is to find the equilibrium concentration of hydroxide ions, \([\text{OH}^-]\).

Setting Up the Calculation: The ICE Table

Tracking Concentration Changes

To determine the equilibrium concentrations, we use an ICE (Initial, Change, Equilibrium) table. Let 'x' represent the change in concentration as the reaction proceeds towards equilibrium. Specifically, 'x' will be the concentration of OH⁻ (and NH₄⁺) formed.

ICE Table for Ammonia Ionization

This table summarizes the concentrations of the species involved in the equilibrium:

Species	Initial Concentration (M)	Change in Concentration (M)	Equilibrium Concentration (M)
NH₃	0.0750	\(-x\)	\(0.0750 - x\)
NH₄⁺	0	\(+x\)	\(x\)
OH⁻	0	\(+x\)	\(x\)

This table provides the terms we need to substitute into the \(K_b\) expression.

Substituting into the \(K_b\) Expression

Now, substitute the equilibrium concentrations from the ICE table into the \(K_b\) expression:

\[ K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]} = \frac{(x)(x)}{(0.0750 - x)} \] \[ 1.75 \times 10^{-5} = \frac{x^2}{0.0750 - x} \]

Deriving \(x^2 = 1.31 \times 10^{-6}\): The Core Computation

Applying the Small x Approximation

Because the \(K_b\) value (\(1.75 \times 10^{-5}\)) is significantly smaller than the initial concentration (0.0750 M), we can often assume that the amount of ammonia that ionizes (x) is negligible compared to the initial concentration. This simplifies the calculation considerably.

The approximation is:

\[ 0.0750 - x \approx 0.0750 \]

This assumption is generally valid if x is less than 5% of the initial concentration. We will verify this later.

Step-by-Step Calculation for \(x^2\)

Applying the approximation to the \(K_b\) expression:

\[ 1.75 \times 10^{-5} \approx \frac{x^2}{0.0750} \]

To find \(x^2\), we rearrange the equation by multiplying both sides by 0.0750:

\[ x^2 \approx (1.75 \times 10^{-5}) \times (0.0750) \]

Now, let's perform the multiplication:

Computation Details:

Multiply the numerical coefficients: \(1.75 \times 0.0750 = 0.13125\)
Include the power of 10: The calculation becomes \(0.13125 \times 10^{-5}\)
Convert to standard scientific notation: To express the coefficient between 1 and 10, move the decimal point one place to the right. This requires decreasing the exponent by 1 (from -5 to -6): \[ 0.13125 \times 10^{-5} = 1.3125 \times 10^{-6} \]
Round to significant figures: Both given values (0.0750 M and \(1.75 \times 10^{-5}\)) have three significant figures. Therefore, we round our result to three significant figures: \[ x^2 \approx 1.31 \times 10^{-6} \]

This detailed computation shows exactly how the intermediate value \(x^2 = 1.31 \times 10^{-6}\) is obtained using the small x approximation.

Calculating the Hydroxide Ion Concentration ([OH⁻])

Solving for x

Now that we have the value for \(x^2\), we can find x, which represents the equilibrium concentration of hydroxide ions, \([\text{OH}^-]\), by taking the square root:

\[ x = \sqrt{x^2} \approx \sqrt{1.31 \times 10^{-6}} \]

Let's calculate the square root:

\[ x \approx \sqrt{1.31} \times \sqrt{10^{-6}} \] \[ x \approx 1.14455... \times 10^{-3} \]

Rounding to three significant figures (consistent with the given data):

\[ x \approx 1.14 \times 10^{-3} \, \text{M} \]

Some conventions might round 1.145 upwards slightly differently based on intermediate steps. Using the unrounded \(x^2 = 1.3125 \times 10^{-6}\):

\[ x = \sqrt{1.3125 \times 10^{-6}} \approx 1.14564... \times 10^{-3} \, \text{M} \]

Rounding this to three significant figures gives:

\[ [\text{OH}^-] = x \approx 1.15 \times 10^{-3} \, \text{M} \]

Therefore, the hydroxide ion concentration in the 0.0750 M NH₃ solution is approximately \(1.15 \times 10^{-3}\) M.

Verifying the Approximation

Was it valid to assume \(0.0750 - x \approx 0.0750\)? We check the percent ionization:

\[ \text{Percent Ionization} = \frac{x}{[\text{NH}_3]_{initial}} \times 100\% \] \[ \text{Percent Ionization} = \frac{1.15 \times 10^{-3}}{0.0750} \times 100\% \approx 0.01533 \times 100\% \approx 1.53\% \]

Since 1.53% is much less than the common threshold of 5%, our approximation was justified and did not introduce significant error.

Visualizing the Calculation Process

Mindmap Overview

This mindmap outlines the key steps involved in solving this weak base equilibrium problem:

mindmap root["Calculate [OH⁻] in 0.0750 M NH₃ (Kb = 1.75e-5)"] id1["Identify Weak Base Equilibrium"] id1a["NH₃ + H₂O ⇌ NH₄⁺ + OH⁻"] id2["Set up ICE Table"] id2a["Initial: [NH₃]=0.0750, [NH₄⁺]=0, [OH⁻]=0"] id2b["Change: -x, +x, +x"] id2c["Equilibrium: 0.0750-x, x, x"] id3["Write Kb Expression"] id3a["Kb = [NH₄⁺][OH⁻] / [NH₃]"] id3b["1.75e-5 = x² / (0.0750 - x)"] id4["Apply Small x Approximation"] id4a["Assume 0.0750 - x ≈ 0.0750"] id4b["1.75e-5 ≈ x² / 0.0750"] id5["Solve for x²"] id5a["x² ≈ 1.75e-5 * 0.0750"] id5b["x² ≈ 0.13125e-5"] id5c["x² ≈ 1.3125e-6"] id5d["Round: x² ≈ 1.31e-6"] id6["Solve for x ([OH⁻])"] id6a["x = √x² ≈ √1.31e-6"] id6b["x ≈ 1.14e-3 M (or 1.15e-3 M)"] id7["Verify Approximation"] id7a["% Ionization = (x / 0.0750) * 100%"] id7b["1.53% < 5% -> Approximation Valid"]

Factors Influencing Weak Base Calculations

Comparative Analysis

This chart highlights the relative importance of different factors when dealing with weak base equilibrium calculations like the one performed for ammonia. A higher score indicates greater significance or impact on the final result or the calculation process itself.

As shown, the correct setup (ICE table, stoichiometry) and the magnitude of \(K_b\) are critically important. The initial concentration and the validity of the approximation significantly influence the calculation pathway and result.

Video Tutorial: Calculating Hydroxide Ion Concentration

Visual Learning Aid

For a broader understanding of calculating hydroxide ion concentrations in various scenarios (including from pH, pOH, Ka, and Kb), the following video provides helpful explanations and examples. While it doesn't specifically cover the exact numbers in your query, it explains the underlying principles and methods used for weak bases like ammonia.

This video tutorial covers methods for finding \([\text{OH}^-]\) using equilibrium constants (\(K_b\)), which is directly relevant to the calculation performed for the ammonia solution.

Frequently Asked Questions (FAQ)

Why is ammonia considered a weak base?

Ammonia (NH₃) is considered a weak base because it only partially ionizes in water. Unlike strong bases (like NaOH) that dissociate completely, only a small fraction of NH₃ molecules react with water to form NH₄⁺ and OH⁻ ions at equilibrium. This limited ionization is reflected in its relatively small base ionization constant (\(K_b = 1.75 \times 10^{-5}\)).

What is the purpose of the 'small x approximation'?

The 'small x approximation' simplifies the calculation of equilibrium concentrations for weak acids and bases. It assumes that the change in concentration (x) due to ionization is negligible compared to the initial concentration. This allows us to avoid solving a quadratic equation (\(K_b = \frac{x^2}{C_{initial} - x}\) becomes \(K_b \approx \frac{x^2}{C_{initial}}\)). The approximation is generally valid when the percent ionization is less than 5%.

When would I need to use the quadratic formula instead of the approximation?

You would need to use the quadratic formula to solve for x (without the approximation) if the 'small x approximation' is not valid. This typically occurs when:

The \(K_a\) or \(K_b\) value is relatively large (e.g., closer to \(10^{-3}\) or \(10^{-4}\)).
The initial concentration of the weak acid or base is very low.

In such cases, the percent ionization exceeds 5%, and ignoring 'x' in the denominator (\(C_{initial} - x\)) introduces significant error. You can always check the validity by calculating the percent ionization after using the approximation; if it's >5%, recalculate using the quadratic formula: \(x^2 + K_b x - K_b C_{initial} = 0\).

How are [OH⁻], pOH, and pH related?

These values are interrelated measures of acidity and basicity in aqueous solutions at 25°C:

pOH: Calculated directly from the hydroxide ion concentration: \( \text{pOH} = -\log_{10}[\text{OH}^-] \)
pH: Related to pOH through the ion product of water (\(K_w = 1.0 \times 10^{-14}\)): \( \text{pH} + \text{pOH} = 14 \)
Therefore, once you calculate \([\text{OH}^-]\), you can find pOH and then pH. For our result \([\text{OH}^-] \approx 1.15 \times 10^{-3}\) M, the pOH \(\approx -\log(1.15 \times 10^{-3}) \approx 2.94\), and the pH \(\approx 14 - 2.94 = 11.06\).