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CHEM PLEASE HELP!!!!?
11.8 mL of 0.3M HCl is added to a 122mL sample of 0.21 M HNO2 (Ka for HNO2=4.0x10^-4). What is the equilibrium concentration of NO2- ions? I have no idea how do this please help.
1 Answer
- FernLv 75 years agoFavorite Answer
Moles HCl = 0.30 moles/liter x 11.8 mL x 1 liter/1000 mL= 0.00354
Volume of HCl after mixing = 11.8 + 122 ml = 133.8 mL
Molarity of HCl after mixing = 0.00354 moles/0.1338 Liters = 0.0265
Same method molarity of HNO2 = 0.191 M
HNO2 <====> H+ + NO2-
[H+][NO2-] / [HNO2] = 4 x 10^-4
initial [H+] from HCl(a strong acid) = 0.0265
………HNO2]…..[H+]…….[NO2-]
initial…0.191…..0.0265………0
change….-x……..+x………….x
equil…0.191-x…0.0265 +x…….x
(0.0265 + x)(x)/ 0.191-x = 4.0 x 10^-4 ;
neglect the x in (0.191 –x) since it is small compared to 0.191
0.0265x + x^2 = 7.64 x 10^-5
x^2 + 0.0265x - 7.64 x 10^-5
solving the quadratic:
x = 0.00262
[NO2-] = 0.00262