February 28, 2017

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!

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