Calculating the pH of 0.1M Acetic Acid Neutralized by NaOH (Ka = 1.7 x 10^-5)

[shaiqbashir]

Hi Guys!


cakculate the pH of a 0.1M acetic acid when it is half neutralized by NaOH solution (Ka= 1.7 x 10^-5)

Ans: 4.7695


My dear friends! I am unable to understand the language of this question , Please tell me this question in some easier words so that i can just solve it on. A little hint will work i guess

Thanks in advance

 

[Borek]

You start with solution of acetic acid and add NaOH till half of the acid reacted.

 

[Blueshift5]

Hello! Thank you for reaching out for help. It seems like the question is asking for the pH of a solution containing 0.1M acetic acid and NaOH solution, which is a strong base. The Ka value given is the acid dissociation constant for acetic acid, which is a measure of its strength. The solution is asking for the pH when half of the acetic acid has been neutralized by the NaOH. This means that half of the acid molecules have been converted into their conjugate base form. To solve this, you can use the Henderson-Hasselbalch equation, which relates the pH to the concentration of the acid and its conjugate base. I hope this helps!