Answer:
1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL
Explanation:
<em>The pKa of the picolinic acid is 5.4.</em>
Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:
pH = pKa + log [Picolinate] / [Picolinic]
<em>Where [] could be taken as moles of each species</em>
<em />
5.61 = 5.4 + log [Picolinate] / [Picolinic]
0.21 = log [Picolinate] / [Picolinic]
1.62181 = [Picolinate] / [Picolinic] <em>(1)</em>
<em></em>
Now, both picolinate and picolinic acid will be:
0.100L * (0.109mol / L) =
0.0109 moles = [Picolinate] + [Picolinic] <em>(2)</em>
<em></em>
First, as we will start with picolinic acid, we need add:
0.0109 moles picolinic acid * (123.10g/mol) = 1.342g of picolinic acid
Now, replacing (2) in (1):
1.62181 = 0.0109 moles - [Picolinic] / [Picolinic]
1.62181 [Picolinic] = 0.0109 moles - [Picolinic]
2.62181 [Picolinic] = 0.0109 moles
[Picolinic] = 4.157x10⁻³ moles
And:
[Picolinate] = 0.0109 - 4.157x10⁻³ moles =
<h3>6.743x10⁻³ moles</h3><h3 />
To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:
Picolinic acid + NaOH → Picolinate + Water
<em>That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH</em>
<em />
6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:
6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =
<h3>6.743mL of the 1.0M NaOH must be added</h3><h3 />
Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:
<h3>Dilute the mixture to 100mL, the volume we need to prepare</h3>
