Respuesta :
Explanation:
Relation between pH and [tex]pK_{a}[/tex] is as follows.
pH = [tex]pK_{a} + log \frac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}[/tex]
Hence, total moles of acetate is calculated as follows.
[tex]0.250 M \times 0.5 L[/tex]
= 0.125 M
[tex]\frac{B}{A} = 10^{{pH} - pk_{a}}[/tex]
= [tex]10^{3.50 - 4.74}[/tex]
= 0.0575
Let us assume that there are x moles of acid. Hence, moles of B are as follows.
B = (0.125 - x) moles
[tex]\frac{0.125 - x}{x}[/tex] = 0.0575
x = 0.118 moles
Now, we will calculate the concentration as follows.
5.25 = 4.74 + [tex]log \frac{[B]}{[A]}[/tex]
[tex]log \frac{[B]}{[A]} = 10^{5.25 - 4.74}[/tex]
= 3.236
x = [tex]\frac{0.125}{4.236}[/tex]
= 0.0295 moles
Therefore, moles of NaOH needed to add are as follows.
[tex]M_{NaOH} \times V_{NaOH}[/tex] = (0.118 - 0.0295) moles
[tex]1.00 M \times V_{NaOH}[/tex] = 0.0885 moles
= [tex]0.0855 L \times \frac{10^{3}ml}{L}[/tex]
= 88.5 ml
Thus, we can conclude that there is 88.5 ml of 1.000 M NaOH must you add in order to change the pH to 5.25.
Volume is the portion of the area occupied by the substance in the object. It is calculated in L, mL, etc. To change pH, 88.5 mL of NaOH must be added.
What is the relation between volume and pH?
An increase in the volume decreases the molarity and for acidic solutions, the pH will increase while for basic the pH will decrease.
The total moles of acetate is the product of the molarity and volume and is 0.125 moles.
The pH and pKa can be shown as,
[tex]\begin{aligned}\rm pH &= \rm pKa + log \dfrac{[CH_{3}COO^{-}]}{[CH_{3}COOH]}\\\\\rm \dfrac{B}{A} &= \rm 10^{pH-pKa}\\\\&=0.0575\end{aligned}[/tex]
Moles of B can be calculated by letting the moles of A = x as,
[tex]\begin{aligned} \rm B &= \rm (0.125 - x) moles\\\\\rm \dfrac {0.125-x}{x}&= 0.0575\\\\\rm x &= 0.118 \;\rm moles\end{aligned}[/tex]
The concentration can be calculated as:
[tex]\begin{aligned}5.25 &= 4.74+\rm log\dfrac{B}{A}\\\\\rm log \dfrac{[B]}{[A]} &= 10^{5.25-4.74}\\\\&= 3.236\end{aligned}[/tex]
The number of the moles (x) = 0.0295 moles.
The volume of sodium hydroxide that can be added is:
[tex]\begin{aligned} \rm M_{NaOH} \times V_{NaOH} &= (0.118 - 0.0295)\\\\1.00 \times \rm V_{NaOH} &= 0.0885\;\rm moles\\\\&= 88.5\;\rm mL\end{aligned}[/tex]
Therefore, 88.5 mL of sodium hydroxide can be added.
Learn more about volume and pH here:
https://brainly.com/question/22559098
