Respuesta :
Answer : The pH at equivalence is, 9.21
Explanation : Given,
Concentration of [tex]HCH_3CO_2[/tex] = 0.5870 M
Volume of [tex]HCH_3CO_2[/tex] = 90.0 mL = 0.09 L (1 L = 1000 mL)
First we have to calculate the moles of [tex]HCH_3CO_2[/tex]
[tex]\text{Moles of }HCH_3CO_2=\text{Concentration of }HCH_3CO_2\times \text{Volume of }HCH_3CO_2[/tex]
[tex]\text{Moles of }HCH_3CO_2=0.5870M\times 0.09L=0.0528mol[/tex]
As we known that at equivalent point, the moles of [tex]HCH_3CO_2[/tex] and NaOH are equal.
So, Moles of NaOH = Moles of [tex]HCH_3CO_2[/tex] = 0.0528 mol
Now we have to calculate the volume of NaOH.
[tex]\text{Volume of }NaOH=\frac{\text{Moles of }NaOH}{\text{Concentration of }NaOH}[/tex]
[tex]\text{Volume of }NaOH=\frac{0.0528mol}{0.4794M}[/tex]
[tex]\text{Volume of }NaOH=0.0253L[/tex]
Total volume of solution = 0.09 L + 0.0253 L = 0.1153 L
Now we have to calculate the concentration of KCN.
The balanced equilibrium reaction will be:
[tex]HCH_3CO_2+NaOH\rightleftharpoons CH_3CO_2Na+H_2O[/tex]
Moles of [tex]CH_3CO_2Na[/tex] = 0.0528 mol
[tex]\text{Concentration of }CH_3CO_2Na=\frac{0.0528mol}{0.1153L}=0.458M[/tex]
At equivalent point,
[tex]pH=\frac{1}{2}[pK_w+pK_a+\log C][/tex]
Given:
[tex]pK_w=14\\\\pK_a=4.76\\\\C=0.458M[/tex]
Now put all the given values in the above expression, we get:
[tex]pH=\frac{1}{2}[14+4.76+\log (0.458)][/tex]
[tex]pH=9.21[/tex]
Therefore, the pH at equivalence is, 9.21
