Question

An aqueous solution contains dissolved and . The concentration of is 0.50 M and pH is 4.20. Calculate the concentration of in this buffer solution.

Answer 1

Here is the correct format for the question.

An aqueous solution contains dissolved
`\mathbf{C_6H_5NH_3Cl}}` and
`\mathbf{C_6H_5NH_2}`. The concentration of
`\mathbf{C_6H_5NH_2}` is 0.50 M and pH is 4.20. Calculate the concentration of
`\mathbf{C_6H_5NH_3^+}` in this buffer solution.
`\mathbf{K_b = 3.8 * 10^(-10)}`

Answer:

`\mathbf{C_6H_5NH_3^+ = 1.1995 M }`

Step-by-step explanation:

From the above information:

`pKb \ of \ C _6H_5NH_2 = -log(3.8 * 10^(-10))`

`pKb \ of \ C _6H_5NH_2 = 9.42`

pH = 4.20

pOH = 14 - pH

pOH = 14 - 4.20 = 9.8

According to Henderson Hasselbalch equation

`pOH = pKb + log ([C_6H_5NH_3^+])/([C_6H_5NH_2])`

`9.8 = 9.42 + log ([C_6H_5NH_3^+])/((0.50))`

`9.8-9.42 = log ([C_6H_5NH_3^+])/((0.50))`

`0.38 = log ([C_6H_5NH_3^+])/((0.50))`

`10^(0.38) = ([C_6H_5NH_3^+])/((0.50))`

`2.399 = ([C_6H_5NH_3^+])/((0.50))`

`C_6H_5NH_3^+ = 2.399 *(0.50)`

`\mathbf{C_6H_5NH_3^+ = 1.1995 M }`