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Answer:
1.342g of picolinic acid and 6.743mL of 1.0M NaOH diluting the mixture to 100.0mL
Explanation:
The pKa of the picolinic acid is 5.4.
Using Henderson-Hasselbalch formula for picolinic-picolinate buffer:
pH = pKa + log [Picolinate] / [Picolinic]
Where [] could be taken as moles of each species
5.61 = 5.4 + log [Picolinate] / [Picolinic]
0.21 = log [Picolinate] / [Picolinic]
1.62181 = [Picolinate] / [Picolinic] (1)
Now, both picolinate and picolinic acid will be:
0.100L * (0.109mol / L) =
0.0109 moles = [Picolinate] + [Picolinic] (2)
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 =
6.743x10⁻³ moles
To obtain these moles of picolinate ion we need to make the reaction of the picolinic acid with NaOH:
Picolinic acid + NaOH → Picolinate + Water
That means to obtain 6.743x10⁻³ moles of picolinate ion we need to add 6.743x10⁻³ moles of NaOH
6.743x10⁻³ moles of NaOH that is 1.0M are, in mL:
6.743x10⁻³ moles * (1L / 1mol) = 6.743x10⁻³L * 1000 =
6.743mL of the 1.0M NaOH must be added
Thus, we obtain the desire moles of picolinate and picolinic acid to obtain the buffer we want, the last step is:
Dilute the mixture to 100mL, the volume we need to prepare
Following are the calculation to the mass of picolinic acid:
Let pKa of picolinic acid [tex]= 5.52[/tex]
form the buffer: [tex]\bold{pH = pKa + \log(\frac{A-}{ HA})}[/tex]
Acid concentration[tex]= 0.109[/tex]
We will require the conjugate, which would be formed by interaction of picolinic acid and NaOH.
calcultion to the need:
[tex]pH = pKa + \log(\frac{A-}{ HA})\\\\5.61 = 5.52 + \log(\frac{A-}{ HA})[/tex]
solve for A- by using antilog
[tex]0.09 = \log(\frac{A-}{ HA})\\\\1.230= \frac{A-}{HA}[/tex]
When
[tex]HA = 0.109\\\\A- = 0.134[/tex]
If [tex]V = 100 ml[/tex]then
[tex]n A- = M\timesV = 0.134 \times 0.1 = 0.0134 \ mol\ of\ A-[/tex]
needed
[tex]HA + NAOH \to H_2O + Na^+ \ and \ A-[/tex]
therefore,
ratios =1:1
we need 0.0134 mol of NaOH
[tex]n = M\times V \\\\V = \frac{0.0134}{1} = 0.0134 \ liter \ of\ NaOH[/tex]
but you also want [tex]0.109 M[/tex] of free picolinic acid so
[tex]n = M\times V = 0.109\times 0.1 = 0.0109\ mol \ of \ Acid[/tex]
Therefore:
[tex]n \ acid = 0.0109 + 0.0134\ mol = 0.0243 \ mol\ of \ acid[/tex]
Preparation:
by add volume (19.07) in NaOH
M = 1.0 to the beaker, and by add water until the beaker marks 1 L
Then add 0.0243 mol of picolinic acid
[tex]\to m = 0.0243 \times 123 = 2.9889\ grams[/tex] of picolinic acid
stir and pH will be that of buffer.
Learn more:
brainly.com/question/17156849
