Consider A Solution That Contains Both C6H5NH2 And C6H5NH3+ ...

Subject ZIP Search Search Find an Online Tutor Now Ask Ask a Question For Free Login Chemistry Consider a solution that contains both C6H5NH2 and C6H5NH3+. Calculate the ratio [C6H5NH2]/[C6H5NH3+] if the solution has the following pH values. (Assume that the solution is at 25°C.)

a- pH = 3.90

b- pH = 4.47

c-pH = 4.58

d- pH = 4.93

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This is a buffer solution made up of a weak base (C6H5NH2) and the conjugate acid (C6H5NH3+). To find the ratio of these species, we can use the Henderson Hasselbalch equation:

pOH = pKb + log [C6H5NH3+] / [C6H5NH2] ... but first we need the pKb for C6H5NH2 (I find 9.42)

(a) pH = 3.90 so pOH = 14 - 3.90 = 10.1

10.1 = 9.42 +log (conj.acid/base)

log (conj.acid/base) = 0.68

[conj.acid]/[base] = 4.79

[C6H5NH2]/[C6H5NH3+] = 0.209

(b) pH = 4.47 so pOH = 9.53

9.53 = 9.42 + log [conj.acid]/[base]

log[conj.acid]/[base] = 0.11

[conj.acid]/[base] = 1.29

[base]/[conj.acid] = 0.775

If you want to do it directly to find base/conj.acid , i.e. [C6H5NH2]/[C6H5NH3] instead of doing it like the above examples which involves taking the reciprocal at the end, you can use the Ka for C6H5NH3+ (which is 4.58) and then use

pH = pKa + log [C6H5NH2] / [C6H5NH3+]

(c) pH = 4.58 = 4.58 + log [C6H5NH2] / [C6H5NH3+]

log [C6H5NH2] / [C6H5NH3+] = 0

[C6H5NH2] / [C6H5NH3+] = 1.0

(d) pH = 4.93 = 4.58 + log [C6H5NH2] / [C6H5NH3+]

log [C6H5NH2] / [C6H5NH3+] = 0.35

[C6H5NH2] / [C6H5NH3+] = 2.24

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