Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
8g of potassium chromate is dissolved in acid to make 100cm3 solution. A dynamic equilibrium occurs,
2CrO42-(aq) + 2H+(aq) ↔ Cr2O72-(aq) + H2O(l)
(i) What is the Kc expression for above eqiulibrium
(ii)calculate pH at which only 1/5 of original amount of chromate ions remain.
(i) kc = [Cr2O72-]1[H2O]0 / [CrO42-]2 [H+1]2 = [Cr2O72-]1 / [CrO42-]2 [H+1]2
There is no [H2O] term in the numerator, as the reaction is carried out in aqueous medium. Active mass of any amount of pure liquid or pure solid is 1. So, [H2O]0 =1
(2)Number of mole of pot. chromate = wt /gram molecular wt
= 8 /194 = 0.04123
From the balanced equation, the No. of mole of H1+ ions required = 0.04123
Concentration of H+1 = No of mole of H+1 / Volume of solution in litres
Concentration of H+1 = 0.04123 / 0.1 = 0.4123 M
pH = log 10 [1 / 0.4123] = log 10 2.425 = 0.3847
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !