1ml 0.1M HCl

99ml 1.4M acetic

What is the pH of the two?

To calculate the pH of a solution, you need to know the concentration of hydrogen ions (H+) in the solution. The concentration of H+ ions determines whether a solution is acidic or basic. The pH scale ranges from 0 to 14, where a pH of 0-6 indicates acidity, 7 indicates neutrality, and 8-14 indicates alkalinity.

To determine the pH of the given solutions, you first need to find the concentration of H+ ions in each solution. In this case, we have two solutions with different concentrations of acids:

Solution 1: 1 mL of 0.1 M HCl
Solution 2: 99 mL of 1.4 M acetic acid

Let's calculate the pH for each solution step by step:

Solution 1: 1 mL of 0.1 M HCl
The concentration of HCl is given as 0.1 M. HCl is a strong acid, which means it fully dissociates in water, releasing one H+ ion for every HCl molecule. Therefore, the concentration of H+ ions in the solution is also 0.1 M.

To calculate the pH, we can use the formula: pH = -log[H+]
Substituting the value of [H+], we get: pH = -log(0.1)
Calculating this using a calculator, the pH of Solution 1 is approximately 1.

Solution 2: 99 mL of 1.4 M acetic acid
Acetic acid (CH3COOH) is a weak acid that partially dissociates in water. To determine the concentration of H+ ions, we need to consider the acid dissociation constant (Ka) of acetic acid, which is approximately 1.8 × 10^-5.

The equation for the dissociation of acetic acid is: CH3COOH ⇌ CH3COO- + H+
Let's assume the concentration of H+ ions at equilibrium is x M.

Using the equilibrium expression for the dissociation reaction, we have: Ka = [CH3COO-][H+] / [CH3COOH]

Given that [CH3COOH] = 1.4 M and [CH3COO-] = x, we can substitute these values into the equation and solve for x.

x = (Ka * [CH3COOH]) / [CH3COO-]
x = (1.8 × 10^-5 * 1.4) / x

Simplifying this equation, we can solve for x:
x^2 = (1.8 × 10^-5 * 1.4)
x = √((1.8 × 10^-5 * 1.4))
x ≈ 0.007

So, the concentration of H+ ions is approximately 0.007 M.

Using the same pH formula, we can calculate the pH for Solution 2: pH = -log(0.007)
Calculating this using a calculator, the pH of Solution 2 is approximately 2.15.

Therefore, the pH of Solution 1 is approximately 1, and the pH of Solution 2 is approximately 2.15.