Question

Which of the following is not true? (a) The conjugate base of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is \(\mathrm{HPO}_{4}^{2-}\). (b) \(\mathrm{pH}+\mathrm{pOH}=14\) for all aqueous solutions. (c) The \(\mathrm{pH}\) of \(1 \times 10^{-8} \mathrm{M} \mathrm{HCl}\) is 8 . (d) 96500 coulombs of electricity when passed through a \(\mathrm{CuSO}_{4}\) solution deposit \(1 \mathrm{~g}\) equivalent of copper at the cathode.

Short answer

Statement (c) is not true.

Step-by-step solution

Understand Conjugate Bases

The conjugate base is formed when an acid donates a proton (^+). For H_2PO_4^-, removing one proton results in HPO_4^{2-}, confirming statement (a) is true.

pH and pOH Relationship

In aqueous solutions at 25°C, the sum of  ext{pH} and  ext{pOH} is always 14, thus statement (b) is generally true under these conditions.

Calculating pH of Dilute HCl Solution

For a solution with [H^+] of 1 imes 10^{-8} ext{ M}, we need to consider the contributions of H extsuperscript{+} ions from water (1 imes 10^{-7} ext{ M}). The real concentration is slightly more than 1 imes 10^{-7} ext{ M}, giving a pH less than 7, contradicting statement (c).

Understanding Electrolysis Depositions

Passing 96500 coulombs (1 Faraday) through a  ext{CuSO}_4 solution deposits one gram equivalent (according to Faraday's laws), making statement (d) true.

Key concepts

Conjugate Base

In chemistry, a conjugate base is what remains after an acid has donated a proton during a chemical reaction. If we take the example of \( \text{H}_2\text{PO}_4^- \), which is dihydrogen phosphate, it can donate a proton to form its conjugate base, \( \text{HPO}_4^{2-} \). This new ion, hydrogen phosphate, is more negatively charged by one unit due to the loss of the proton, which carried a positive charge.
Understanding conjugate bases is crucial because they help in predicting the behavior of chemical species in different reactions. When an acid loses a proton, its conjugate base can potentially regain a proton in the reverse reaction, showcasing the dynamic nature of acid-base interactions.
Some people may find it hard to visualize these transformations, but remember that these are about the transfer of protons, not the molecule's basic structure.

pH and pOH Relationship

The relationship between pH and pOH in aqueous solutions is fundamental in understanding acidity and alkalinity. At 25°C, the sum of pH and pOH is always equal to 14. This relation is derived from the ion product of water, \( K_w = [\text{H}^+][\text{OH}^-] = 1 \times 10^{-14} \), and taking the negative logarithm of both sides gives this constant sum.

Here's a breakdown of why this relationship matters:

• **Neutral Solution**: pH = 7 and pOH = 7. Both hydrogen and hydroxide ion concentrations are equal.
• **Acidic Solution**: pH < 7, meaning the solution has more \( \text{H}^+ \) ions.
• **Basic Solution**: pH > 7, indicating more \( \text{OH}^- \) ions.

This rule of thumb simplifies understanding complex chemical environments by giving a straightforward reference point.
One common mistake is ignoring the condition of temperature. While the pH + pOH = 14 rule holds at 25°C, deviations occur at different temperatures due to changes in \( K_w \).

Electrolysis

Electrolysis is a process that involves using electrical energy to drive a non-spontaneous chemical reaction. In the case of \( \text{CuSO}_4 \) (Copper Sulfate) solution, electricity can be passed through the solution to deposit copper metal at the cathode, while oxygen is released at the anode.
Let's break down the concepts related to this process:

• **Faraday's Laws of Electrolysis**: One faraday of electricity (96500 coulombs) is required to deposit one mole of electrons on a substance. This means you can calculate the amount of substance deposited if you know the charge passed.
• **Cathode and Anode**: During electrolysis of \( \text{CuSO}_4 \), copper ions (\( \text{Cu}^{2+} \)) gain electrons at the cathode to form copper metal. Meanwhile, water at the anode discharges oxygen gas.

Understanding electrolysis is vital in industries that involve metal extraction and refining. The calculated electrochemical equivalents from Faraday's laws ensure efficient processing, such as determining the appropriate quantity of copper metal from a given amount of charge. Remember, each step in electrolysis is crucial for the complete picture of how energy and matter interact.