Question

Use the Henderson-Hasselbach equation to calculate the pH of each solution:

I figured out part A by myself because the numbers were already in Molarity for me so I just plugged in numbers and calculated.

B. A solution that contains 0.785% C5H5N by mass and 0.985% C5H5NHCl by mass. (Here, I don't know what to do with the %'s to get the molarity in order to find the pH.)

C. A solution that contains 15.0 g of HF and 25.0 g of NaF in 125 mL of solution. (Here,i thought i could just use the g and mL to find molarity using mol/L but that led me to the wrong answers.)

Please explain your process clearly as possible because I would really like to be prepared when i see these kinds of questions on my exam. Thanks!

Answer 1

B. )

0.785% C5H5N means 0.785 g present in 100 g of solution that means solute mass 0.785 g

moles of C5H5N = mass / molar mass

                           = 0.785 / 79.09

                           = 9.925 x 10^-3

moles of C5H5NHCl = 0.985 / 115.6

                                = 8.52 x 10^-3

pKb of C5H5N = 8.77

pOH = pKb + log [C5H5NHCl / C5H5N]

        = 8.77 + log (8.52 x 10^-3 / 9.925 x 10^-3)

       = 8.70

pH + pOH = 14

pH = 5.30

C) moles of HF = 15 / 20 = 0.75

   moles of NaF = 25 / 42 = 0.595

HF pKa = 3.20

pH = pKa + log [NaF / HF]

pH = 3.20 + log (0.595 / 0.75)

pH = 3.10