Email
Chat with tutor
logo

Ask Questions, Get Answers

X
 
Questions  >>  JEEMAIN and NEET  >>  Chemistry  >>  Equilibrium
Answer
Comment
Share
Q)

Degree of dissociation of an acid HCl is 95%. 0.192 g of the acid is present in 0.5 L of solution. The pH of the solution is

(a) 2

(b) 1

(c) 3

(d) 4

1 Answer

Comment
A)
Toolbox:
  • $\rho H = - log [H^+]$
 
Answer: 2
 
No. of moles of $HCl = \frac{0.192}{36.5}$ = 0.0053
Volume = 0.5 litre
 
Molarity of $HCL = \frac{0.0053}{0.5} = 1.06 \times 10^{-2}$ M
 
$[H^+]$ in $HCL$ solution = $1.06 \times 10^{-2} \times \frac{95}{100} = 100.7 \times 10^{-4} \approx 1 \times 10^{-2} $
 
Thus, $\rho H = - log [H^+] = - log (1 \times 10^{-2}) = 2$

 

Help Clay6 to be free
Clay6 needs your help to survive. We have roughly 7 lakh students visiting us monthly. We want to keep our services free and improve with prompt help and advanced solutions by adding more teachers and infrastructure.

A small donation from you will help us reach that goal faster. Talk to your parents, teachers and school and spread the word about clay6. You can pay online or send a cheque.

Thanks for your support.
Continue
Please choose your payment mode to continue
Home Ask Homework Questions
Your payment for is successful.
Continue
Clay6 tutors use Telegram* chat app to help students with their questions and doubts.
Do you have the Telegram chat app installed?
Already installed Install now
*Telegram is a chat app like WhatsApp / Facebook Messenger / Skype.
...