Two solutions A and B each of 100L was made by dissolving 4g of NaOH and 9.8g of H2SO4 in water, respectively. The pH of the resultant solution obtained from mixing 40L of solution A and 10L of solution B is_________:
Option: 1 pH = 10.6
Option: 2 pH = 12.7
Option: 3 pH = 9.8
Option: 4 pH = 11
 

Answers (1)

For given solutions, we have:

Moles of NaOH  4/40 = 0.1 moles

Moles of H2SO4 = 9.8/98 = 0.1 moles

Molarity of NaOH = 0.1/100L
And molarity of H2SO4 = 0.1/100

Now, 40L of NaOH solution and 10L of H2SO4 solution are added, thus we get:

Total volume = 50L

Milliequivalents of NaOH = 40x(0.1/100)x1 = 0.04

Milliequivalents of H2SO4 = 10x(0.2/100)x2 = 0.02

Thus, Meq of NaOH left = 0.04 - 0.02 = 0.02

[OH-] = 4x10-4

pOH = -log[4x10-4]

pOH = -log4 - log10-4

pOH = -0.60 + 4 = 3.4

Further, we know:

pH = 14 - 3.4

pH = 10.6

Hence, the option number (1) is correct.

Most Viewed Questions

Preparation Products

Knockout JEE Main (Six Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 9999/- ₹ 8499/-
Buy Now
Knockout JEE Main (Nine Month Subscription)

- AI Coach Study Modules, - Unlimited Mock Tests, - Study Improvement Plan.

₹ 13999/- ₹ 12499/-
Buy Now
Test Series JEE Main 2024

Chapter/Subject/Full Mock Tests for JEE Main, Personalized Performance Report, Weakness Sheet, Complete Answer Key,.

₹ 7999/- ₹ 4999/-
Buy Now
JEE Main Rank Booster 2023

Booster and Kadha Video Lectures, Unlimited Full Mock Test, Adaptive Time Table, Faculty Support.

₹ 9999/- ₹ 6999/-
Buy Now
Knockout JEE Main (One Month Subscription)

Personalized AI Tutor and Adaptive Time Table, Self Study Material, Unlimited Mock Tests and Personalized Analysis Reports, Faculty Support.

₹ 7999/- ₹ 4999/-
Buy Now
Boost your Preparation for JEE Main 2021 with Personlized Coaching
 
Exams
Articles
Questions