Asked by Mary
What is the (H30) concentration when 40 mL of 0.10 M nitrous acis, HNO2, are added to 10.0 mL of 0.10 M HCl? Ka for HNO2 is 4.0 x 10^-4.
I get 1 x 10^-3. Is this right?
I get 1 x 10^-3. Is this right?
Answers
Answered by
DrBob222
I didn't get your answer.
millimoles HCl = 10.0 x 0.1 = 1 mmole.
(HCl) = 1 mmole/50 mL = 0.02 M
HNO2 ==> H^+ + NO2^-
The H^+ from HCl is a common ion here; therefore, the equilibrium for HNO2 is shifted to the left.
Ka = (H^+)(NO2^-)/(HNO2)
Substitute 0.02 for (H^+). Calculate (HNO2) = 40 x 0.1 = 4 millimoles and that divided by 50 mL = 0.08 M. Then add H^+ from HCl and NO2^- (that will be the equivalent H^+ from the HNO2) to obtain total H^+. Something like 0.022 or close to that.
millimoles HCl = 10.0 x 0.1 = 1 mmole.
(HCl) = 1 mmole/50 mL = 0.02 M
HNO2 ==> H^+ + NO2^-
The H^+ from HCl is a common ion here; therefore, the equilibrium for HNO2 is shifted to the left.
Ka = (H^+)(NO2^-)/(HNO2)
Substitute 0.02 for (H^+). Calculate (HNO2) = 40 x 0.1 = 4 millimoles and that divided by 50 mL = 0.08 M. Then add H^+ from HCl and NO2^- (that will be the equivalent H^+ from the HNO2) to obtain total H^+. Something like 0.022 or close to that.
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