A chemist titrates 90.0 mL of a 0.5870 M acetic acid (HCH3CO2) solution with 0.4794M NaOH solution at 25 °C. Calculate the pH at equivalence. The p Kg of acetic acid is 4.76. Round your answer to 2 decimal places.

Answer : The pH at equivalence is, 9.21

Explanation : Given,

Concentration of = 0.5870 M

Volume of = 90.0 mL = 0.09 L (1 L = 1000 mL)

First we have to calculate the moles of

As we known that at equivalent point, the moles of and NaOH are equal.

So, Moles of NaOH = Moles of = 0.0528 mol

Now we have to calculate the volume of NaOH.

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:

Moles of = 0.0528 mol

At equivalent point,

Given:

Now put all the given values in the above expression, we get:

Therefore, the pH at equivalence is, 9.21

sorry i was late, but i don't know what assignment your talking about, i'm in flvs as well, im in earth space science. but u are behind pace bc every class should be in the 3's or 4'