Calculate the pH resulting from mixing 15.00 mL of 0.800 M HIO3 (pKa = 0.77) with 45.00 mL of 0.0200 M NaOH.

pH = -0.32

Explanation:

The reaction of HIO₃ with NaOH is:

HIO₃ + NaOH → IO₃⁻ + Na⁺ + H₂O

Moles that react are:

15.00mL = 0.015L * (0.800mol / L) = 0.012 moles HIO₃

45.00mL = 0.045L * (0.0200 mol / L) = 9x10⁻⁴ moles NaOH.

The moles of NaOH that reacts are equal to moles of IO₃⁻, and moles of HIO₃ are:

0.012 mol - 9x10⁻⁴ moles = 0.0111 moles HIO₃

Now, the mixture of a weak acid, HIO₃ and IO₃⁻, its conjugate base, produce a buffer. The pH of the buffer is obtained using H-H equation:

pH = pKA + log [IO₃⁻] / [HIO₃]

Where [] could be taken as moles of each compound.

Replacing:

pH = 0.77 + log [9x10⁻⁴ moles IO₃⁻] / [0.0111 moles HIO₃]

Yes, it is possible to have negative pH's!

introduction of a drug directly into the respiratory tract is a form of inhalation administration.

6.4 cm (max) - 25 cm (min)