Question

Ive been getting the wrong answers when calcuating these. Please help

21.50 mL of 0.320 M NaOH was added to 20.00 mL of 0.410 M HCI. The calculated pH is (2 dec places) Thus the concentration of hydronium is (2 sig figs) Times 10Caret (integer) And the concentration of hydroxide is (2 sig figs) Times 10Caret (integer) In change from the lab manual, rather than pouring the excess NaOH and titration mixtures down the drain at the end of the period, you will pour them into a waste jar. This due to the fact that the pH of the solutions will be greater than 11 and we shouldn't pour things down the drain that have a pH outside of the range of 4-10. Should we pour 0.00100 M HCI down the drain? (yes/no)

Answer 1

4)

Molarity of mixer = (20*0.41+21.5*0.32)/41.5

   = 0.36 M

in the mixture , HCl is excess.

so that

pH = -log(0.36) = 0.44

[H+] = 0.36 M

[OH-] = 10^(-14)/0.36 = 2.78*10^-14 M

5)

pH of HCl = -log0.001 = 3

as the pH is less than 4. we sholud not pour the things down the drain