When a small amount of acid is added to a non-buffered solution, there is a large change in pH. Calculate the pH when 26.9 mL of 0.0044 M HCl is added to 100.0 mL of pure water. Comment and hint in the general feedback.

The correct answer is pH= 2.93

Explanation:

HCl is a strong acid. That means it dissociates completely in water as follows:

HCl (aq) → H⁺(aq) + Cl⁻(aq)

As the dissociation is complete, the amount of H⁺ generated is equal to the concentration of HCl. So, we multiply the volume of HCl added (26.9 mL) by the concentration of HCl (0.0044 M) and then we divide into the volume of water (100.0 mL). Notice that the factor 26.9 mL/100 mL is the dilution factor:

[HCl]= 26.9 mL/100 mL x 0.0044 M = 1.18 x 10⁻³ M

We know that pH= -log [H⁺] and that in this case [H⁺]= [HCl] (because HCl is a strong acid), so we directly calculate the pH from [HCl] as follows:

pH= -log [H⁺]= -log [HCl] = -log (1.18 x 10⁻³ M) = 2.92

Notice that 2.92 is a low value of pH. Hence, when we add a strong acid to water (without a buffer solution), the pH decreases abruptly.

20 drops hope this