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Expert Solution

1. 0.170 M in HNO2 and 9.0×10−2 M in HNO3

Ka of HNO2 = 5.6 x 10^-4

concentration = 0.170 M

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

0.170 0 0

0.170 - x x x

Ka = x^2 / 0.170 - x

5.6 x 10^-4 = x^2 / 0.170 - x

x = 9.48 x 10^-3

[H+] from HNO2 = 9.48 x 10^-3 M

[H+] from HNO3 = 9.0 x 10^-2 M

total [H+] = 0.09948 M

pH = -log [H+] = -log (0.09948)

pH = 1.0

queen_honey_blossom answered 2 years ago

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