Calculating the pH of a 10^-10 M NaOH Solution: A Detailed Guide

Introduction

In this detailed guide, we will walk through the process of calculating the pH of a 10^-10 M sodium hydroxide (NaOH) solution. We'll explore the basic principles of pH calculation and the significant role the ionic product of water (Kw) plays in determining the pH of a basic solution.

Understanding pH and Basic Solutions

The pH of a solution is defined as the negative logarithm of the hydrogen ion concentration [H ] or pH -log[H ]. In a basic solution like NaOH, the concentration of hydroxide ions [OH-] is much higher compared to the concentration of hydrogen ions [H ]. This relationship is crucial in calculating the pH of such a solution.

Calculation Steps

To find the pH of a 10^-10 M NaOH solution, follow these steps:

Step 1: Determine the Concentration of Hydroxide Ions [OH-]

Given the concentration of NaOH, [OH-] is equal to the concentration of NaOH itself. Therefore, for a 10^-10 M NaOH solution:

[OH-] 10^-10 M

Step 2: Calculate the Hydrogen Ion Concentration [H ]

The relationship between the concentrations of hydrogen ions [H ], hydroxide ions [OH-], and the ionic product of water (Kw) is given by:

[H ] Kw / [OH-]

At 25°C, the ionic product of water, Kw, is approximately 1 × 10^-14. Therefore, substituting the values, we get:

[H ] (1 × 10^-14) / (10^-10) 10^-4 M

Step 3: Calculate the pH

The pH is determined by taking the negative logarithm of the hydrogen ion concentration [H ]. Thus:

pH -log[H ] -log(10^-4) 4

Therefore, the approximate pH of a 10^-10 M NaOH solution is 4.

Deceptive Nature of the Problem

This problem is often used as a deceptive exercise in stoichiometry. Students frequently arrive at the incorrect conclusion that a 10^-10 M NaOH solution would have a pH of 4 due to a misunderstanding of logarithmic scales. However, a solution of NaOH can be considered basic, meaning its pH would be greater than 7. We will correct this calculation using the exact method.

Exact Calculation

The exact calculation involves solving the equation [H ][OH-] Kw, which gives:

([H ])(10^-10) 10^-14

Solving for [H ], we get:

[H ] (1 × 10^-14) / (10^-10) 9.99500125 × 10^-8 M

Hence, the pH is:

pH -log(9.99500125 × 10^-8) ≈ 7.000217147

This value falls within the expected range of 7.0001-7.0003.

Additional Insights

The question can be quickly resolved by understanding that the maximum pH possible is 14 (from water’s autoprotolysis), which corresponds to [OH-] 10^-4 M and [H ] 10^-10 M. Thus:

pH pOH 14

Since pOH -log(10^-10) 10, pH must be 14 - 10 4

Conclusion

Understanding how to accurately calculate the pH of a 10^-10 M NaOH solution is crucial for students and professionals alike. It involves recognizing the relationship between [H ], [OH-], and the ionic product of water (Kw) and using logarithmic properties effectively. This guide ensures a clear and comprehensive approach to solving such problems.