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.
3 answers
oops..I meant NaOH is a strong base....
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.
what does ICE stand for? thanks.
2 answers