Respuesta :
Answer : The pH at equivalence is, 9.08
Explanation : Given,
Concentration of [tex]HC_2H_5CO_2[/tex] = 0.1917 M
Volume of [tex]HC_2H_5CO_2[/tex] = 220.0 mL = 0.220 L (1 L = 1000 mL)
First we have to calculate the moles of [tex]HC_2H_5CO_2[/tex]
[tex]\text{Moles of }HC_2H_5CO_2=\text{Concentration of }HC_2H_5CO_2\times \text{Volume of }HC_2H_5CO_2[/tex]
[tex]\text{Moles of }HC_2H_5CO_2=0.1917M\times 0.220L=0.0422[/tex]
As we known that at equivalent point, the moles of [tex]HC_2H_5CO_2[/tex] and KOH are equal.
So, Moles of KOH = Moles of [tex]HC_2H_5CO_2[/tex] = 0.0422 mol
Now we have to calculate the volume of KOH.
[tex]\text{Volume of }KOH=\frac{\text{Moles of }KOH}{\text{Concentration of }KOH}[/tex]
[tex]\text{Volume of }KOH=\frac{0.0422mol}{0.1787M}[/tex]
[tex]\text{Volume of }KOH=0.00754[/tex]
Total volume of solution = 0.220 L + 0.00754 L = 0.22754 L
Now we have to calculate the concentration of KCN.
The balanced equilibrium reaction will be:
[tex]HC_2H_5CO_2+KOH\rightleftharpoons C_2H_5CO_2K+H_2O[/tex]
Moles of [tex]C_2H_5CO_2K[/tex] = 0.0422 mol
[tex]\text{Concentration of }C_2H_5CO_2K=\frac{0.0422mol}{0.22754L}=0.1855M[/tex]
At equivalent point,
[tex]pH=\frac{1}{2}[pK_w+pK_a+\log C][/tex]
Given:
[tex]pK_w=14\\\\pK_a=4.89\\\\C=0.1855M[/tex]
Now put all the given values in the above expression, we get:
[tex]pH=\frac{1}{2}[14+4.89+\log (0.1855)][/tex]
[tex]pH=9.08[/tex]
Therefore, the pH at equivalence is, 9.08
