Posted by lisa on Monday, January 14, 2013 at 8:33pm.
Give the data clearly.
so this is the question:
you will be given a sample of nictinic acid with an unknown concentration, and it will be titrated with 0.100M NaOH. instead of measuring the volume of the basic solution required to reach the endpoint, however, you will measure the pH. Then you will graph the data you collected and use the graph to calculate the molar concentration of the nicotinic acid and its ka.
the data is
Vol. of NaOH added
0
1
2
3
4
5
10
15
20
21
22
23
24
25
26
27
28
29
30
35
40
45
50
pH
2.92
3.47
3.79
3.98
4.13
4.25
4.67
5.03
5.57
5.72
5.91
6.23
8.78
11.29
11.59
11.75
11.87
11.96
12.22
12.36
12.46
12.52
this is the data and this was given
OK. You will add the 0.1M NaOH, read the volume NaOH and the pH for each addition. I can't graph it but I will post a link that will show you how it should look. Then you will use the graph to determine the volume of NaOH used to reach the equivalence point and from there the concn of the nicotinic acid. After you know the equivalence point, you take 1/2 that volume, read the pH at the half-way point, and that will be pKa. You get Ka from pKa.
The curve will look something like this.
http://www.google.com/imgres?hl=en&client=firefox-a&hs=Szx&sa=X&tbo=d&rls=org.mozilla:en-US:official&tbm=isch&tbnid=YMzuN-RiHqx1NM:&imgrefurl=http://www.titrations.info/titration-basic-terms&docid=N4qxMJfzM-kjoM&imgurl=http://www.titrations.info/img/acetic-acid-titration-curve.png&w=368&h=284&ei=XcT0UIbvO6KE2wXszoCoDg&zoom=1&iact=hc&vpx=139&vpy=122&dur=13504&hovh=197&hovw=256&tx=149&ty=90&sig=111854003357503651221&page=2&tbnh=140&tbnw=194&start=8&ndsp=12&ved=1t:429,r:13,s:0,i:137&biw=846&bih=468
i know how to grapgh it and all i am only stuck with how to find the intial molar concentration. i believe you're telling me the step to finding the ka value.
sorry i am a little confused with this
thanks for the help i really appreciate it :D
The equivalence point is between 25 and 26 mL. The question you posted asks for Ka as well as molar concn. I told you how to do both.
okay thank you very much :)
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