Asked by Holly
Calculate the pH at the point in the titration of 25.00 mL 0.108 M HNO2 at which 10.00 mL 0.162 M NaOH have been added. For HNO2, Ka=5.1 x 10^-4 and:
HNO2 + OH- -----> H2O + NO2-
I know that HNO2 is a weak acid and that 5.1 x 10^-4=[NO2-]/[HNO2][OH-]. NaOH is a strong acid . Not sure how to go from here. Thanks.
HNO2 + OH- -----> H2O + NO2-
I know that HNO2 is a weak acid and that 5.1 x 10^-4=[NO2-]/[HNO2][OH-]. NaOH is a strong acid . Not sure how to go from here. Thanks.
Answers
Answered by
Holly
oops..I meant NaOH is a strong base....
Answered by
DrBob222
HNO2 + NaOH ==> NaNO2 + H2O
moles HNO2 = M x L = 0.025 x 0.108 = 0.00270 moles HNO2 initially.
moles NaOH added = 10mL x 0.162 = 0.00162
moles HNO2 remaining after reacction = 0.00270-0.00162 =0.00108.
M HNO2 = moles/L (L = 25 mL + 10 mL)
So this problem becomes determine the pH of a 0.00108/0.035)M soln of HNO2. Set up an ICE chart, solve for H^+ and convert to pH.
moles HNO2 = M x L = 0.025 x 0.108 = 0.00270 moles HNO2 initially.
moles NaOH added = 10mL x 0.162 = 0.00162
moles HNO2 remaining after reacction = 0.00270-0.00162 =0.00108.
M HNO2 = moles/L (L = 25 mL + 10 mL)
So this problem becomes determine the pH of a 0.00108/0.035)M soln of HNO2. Set up an ICE chart, solve for H^+ and convert to pH.
Answered by
Holly
what does ICE stand for? thanks.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.