Question

1)Calculate the pH of a 6.50×10^-3M solution of hydroiodic acid, HI.

2)Calculate the pH of a 0.725M solution of nitrous acid, HNO₂.

Answer 1

1) HI is strong so it is completely dissociating

[H+] = 6.50×10^-3M

pH = -log [H+]

pH = - log (6.50×10^-3)

pH = 2.19

2)

HNO2 is the weak acid . we need to use ICE table

HNO2 --------------------> H+   +   NO2-

0.725                              0            0 --------------> I

-x                                  +x            +x --------------> C

0.725-x                           x            x ----------------->E

Ka = [H+][NO2-] / [HNO2]

5.6 x 10^-4 = x^2 / 0.725 -x

x^2 + 5.6 x 10^-4 x - 4.06 x 10^-4 = 0

x = 0.0199

[H+] = 0.0199 M

pH = -log [H+]

pH = -log (0.0199)

pH = 1.70