Textbook Problem 6D.9

[Saiya Smith]

In textbook problem 6D.9, it asks us to find the pH of a solution given the deprotonation percentage. In the ice table, the answer key labeled the concentration of H30+ as x, but when solving for the pH, took the negative log of 2x instead (like instead of taking pH=-log(2.4x0.11) they took pH=-log(2.6x10^-3) which is double the first amount. Why did they do this if x was supposed to represent the concentration of hydronium? Attached is the question and solution for reference.

[Olivia DeCecco 1B]

When solving for the concentration of H3O+ the equation would be [H3O+]/ 0.110 = 2.4x10^-2. Therefore isolating the variable you would get [H3O+] would equal to 2.64x10^-3. This making the pH = -log [2.64x10^-3].