What is the #H_3O^+# and the pH of a buffer that consists of 0.24 M #HNO_2# and 0.68 M #KNO_2#? (#K_a# of #HNO_2 = 7.1 x 10^-4#)?

1 Answer
Jan 3, 2018

Answer:

Explanation:

...which says that...

#pH=pK_a+log_10{[[A^-]]/[[HA]]}#

and so here....#pH=3.15+log_10((0.68*mol*L^-1)/(0.24*mol*L^-1))=3.15+0.453=3.60#

And so.....

#[H_3O^+]=10^(-3.60)*mol*L^-1=2.51xx10^-4*mol*L^-1#.

Solution #pH# is raised relative to #pK_a# given that the concentration of the conjugate base, the nitrite anion, is GREATER than that of the nitrous acid.

What would #pH# be, should #[NO_2^(-)]=[HNO_2]#?