🤓 Based on our data, we think this question is relevant for Professor Fatima's class at UWATERLOO.

For this, we shall use the **Henderson-Hasselbalch equation** as in:

$\overline{){\mathbf{pOH}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{{\mathbf{pK}}}_{{\mathbf{b}}}{\mathbf{}}{\mathbf{+}}{\mathbf{}}{\mathbf{log}}\left(\frac{\mathbf{conjugate}\mathbf{}\mathbf{acid}}{\mathbf{weak}\mathbf{}\mathbf{base}}\right)}$

**C**_{6}H_{5}NH_{2}**is a weak base**→ According to Bronsted-Lowry definition, a base is a proton (H^{+}) acceptor.**C**→_{6}H_{5}NH_{3}^{+ }is the conjugate acidof a base is simply the compound formed when a base gains a hydronium ion (H*conjugate acid*^{+}). It's considered as an acid because it can lose H^{+}to reform the base.

We can get pOH from pH and pK_{b} from K_{b}.

$\overline{){\mathbf{pOH}}{\mathbf{}}{\mathbf{=}}{\mathbf{}}{\mathbf{14}}{\mathbf{-}}{\mathbf{pH}}}\phantom{\rule{0ex}{0ex}}\mathbf{pOH}\mathbf{}\mathbf{=}\mathbf{}\mathbf{14}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{20}$

**pOH = 9.80**

$\overline{){{\mathbf{pK}}}_{{\mathbf{b}}}{\mathbf{}}{\mathbf{=}}{\mathbf{-}}{\mathbf{log}}{\mathbf{}}{{\mathbf{K}}}_{{\mathbf{b}}}}\phantom{\rule{0ex}{0ex}}{\mathbf{pK}}_{{\mathbf{b}}}\mathbf{}\mathbf{=}\mathbf{-}\mathbf{log}\mathbf{}\left(\mathbf{3}\mathbf{.}\mathbf{8}\mathbf{\times}{\mathbf{10}}^{\mathbf{-}\mathbf{10}}\right)$

**pK _{b} = 9.42**

Substituting in the equation:

An aqueous solution contains dissolved C_{6}H_{5}NH_{3}Cl and C_{6}H_{5}NH_{2}. The concentration of C_{6}H_{5}NH_{2} is 0.50 M and pH is 4.20.

a. Calculate the concentration of C_{6}H_{5}NH_{3}^{+} in this buffer solution.

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Based on our data, we think this problem is relevant for Professor Fatima's class at UWATERLOO.

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Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl 2nd Edition practice problems.