Question

With explanation,Calculate the pH of a 0.1M acetic acid when it is half neutralised by NaOH solution (Ka =1.7×10–5).​

Answer 1

Answer:

Here we need volume of acid CH3COOH and base NaOH. If volumes are not known we can't find pH of the resultant solutions.

If we take both acids and bases are of equal volumes of V each. So the solution act as the salt solution for finding pH we need to apply salt hydrolysis formula for finding pH.

Now the mixture will acts as a salt CH3COO-Na+ of week acid CH3COOH and strong base NaOH.

Then pH is 7 + 0.5*(pKa+ logX) here X is concentration of salt.

We known that pKa of CH3COOH is 4.75

Given concentration of acid is 0.1M and base is 0.1 M

Now both are equal volumes and concentrations.

Let's assume concentration of salt is X

Now 2VX = V(0.1+0.1)

X = 0.2/2

X= 0.1

Now pH = 7 + 0.5*(pKa+ logX)

pH = 7 + 0.5*(4.75 + log 0.1)

= 7 + 0.5*(4.75–1)

= 7 + 0.5*(3.75)

= 7 + 1.875

= 8.875.

Therefore pH of the resultant solution is 8.875

Explanation: