Homework Help: chemistry

Posted by Holly on Tuesday, October 4, 2011 at 9:24pm.

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 NaOH is a strong base.

Help given to me so far from Dr. Bob222:
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 reaction = 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……
I set up the ICE chart:
HNO2 + H20   H3O^+ + NO2^- Ka=5.1 x10^-4
I: 3.085x10^-2 M --- ----
C: -X +X +X
E: 3.085x10^-2 –X M XM XM
Since the Ka and the Molarity of the acid are within 10^2, I cannot drop the –X. So, 5.1x10^-4= [X^2]/[3.085x10^-2 –X]. I get a quadratic equation of X^2+5.1x10^-4X -1.5734x10^-5=0. Therefore X=[H3O^+]=0.003719. pH=-log [X]. pH=2.43. My book says the answer is 3.47. I am confused. Please tell me what I am doing wrong. Thank you!

chemistry - DrBob222, Tuesday, October 4, 2011 at 10:12pm
First, I divided 2.70 mmoles (0.00270 moles) by 35 = 0.03086M which isn't enough difference to worry about. Besides we don't have that many sig digits, anyway. The main thing is that you didn't include the NaNO2 that is formed. That is the salt of the weak acid so it is a buffered solution. (NaNO2) = 2.70mmoles/35mL = 0.0771M
Ka = 5.1E-4 = ((H^)(NO2^-)/(HNO2)
5.1E-4 = (x)(x+0.0771)/(0.03086-x).
I worked it both with a quadratic and without and it doesn't make much difference; however, I did NOT obtain 3.47. I found 2.02E-4 for (H^+) for pH = 3.69. I also used the Henderson-Hasselbalch equation which is
pH = pKa + log (NO2^-)/(HNO2)
pH = 3.29 + log(2.70mmols/1.08 mmols)
pH = 3.687 which would round to 3.69
I don't believe the 3.47 answer is correct. Check my work carefully.
chemistry - Holly, Tuesday, October 4, 2011 at 11:30pm
I got it! You took the wrong mol of No2- from the above neutralization rxn....it should be 0.00162 mol NO2-/0.035L=0.0463M NO2-....I did all the calcs and I get 3.47 for pH which is book answer! Thank you for pointing me in the right direction! :-)

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