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.

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.