×

#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-1023-196

+91-120-4616500

CART 0

• 0

MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
```
Calculate the pH of a solution obtained by mixing 0.1 litre of pH = 4 and 0.2 litre of pH = 10

```
3 years ago

## Answers : (1)

Arun
25768 Points
```							 ph of strong acid = 4 [H+] = 10-4 = molarity of acid in 0.1L , moles of acid = 0.1* 10-4 ph of base =10 [H+]=10-10 [OH-]=10-14/10-10=10-4 = molarity of base in 0.2L , moles of acid = 0.2*10-4 now nutralization will occur and acid is completely nutralised by base , at the end base remaining due to higher concentration.... now final molariry = remaining moles/total vol = (0.2*10-4-0.1*10-4 ) /0.3                      concentration of base               =10-4/3 [H+] = 3*10-14/10-4 =3*10-10 ph=-loh[H+] = 10-log3= 9.523
```
3 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

## Other Related Questions on Inorganic Chemistry

View all Questions »

### Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution

### Course Features

• 54 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions

Have any Question? Ask Experts

Post Question