Solution Found!

A 1.475-g sample containing NH4Cl (FM 53.492), K2CO3 (FM 138.21), and inert ingredients

Chapter 26, Problem 26-21

(choose chapter or problem)

Get Unlimited Answers
QUESTION:

A 1.475-g sample containing NH4Cl (FM 53.492), K2CO3 (FM 138.21), and inert ingredients was dissolved to give 0.100 L of solution. A 25.0-mL aliquot was acidified and treated with excess sodium tetraphenylborate, NaB(C6H5)4 , to precipitate K and NH4 ions completely: FM 358.33 FM 337.27 The resulting precipitate amounted to 0.617 g. A fresh 50.0-mL aliquot of the original solution was made alkaline and heated to drive off all the NH3: It was then acidified and treated with sodium tetraphenylborate to give 0.554 g of precipitate. Find the weight percent of NH4Cl and K2CO3 in the original solid. 26-

Questions & Answers

QUESTION:

A 1.475-g sample containing NH4Cl (FM 53.492), K2CO3 (FM 138.21), and inert ingredients was dissolved to give 0.100 L of solution. A 25.0-mL aliquot was acidified and treated with excess sodium tetraphenylborate, NaB(C6H5)4 , to precipitate K and NH4 ions completely: FM 358.33 FM 337.27 The resulting precipitate amounted to 0.617 g. A fresh 50.0-mL aliquot of the original solution was made alkaline and heated to drive off all the NH3: It was then acidified and treated with sodium tetraphenylborate to give 0.554 g of precipitate. Find the weight percent of NH4Cl and K2CO3 in the original solid. 26-

ANSWER:
1.2

The mass of NH4Cl in the original solid = (0.617g/358.33)*53.492 = 0.046 g

The mass of K2CO3 in

Add to cart


Study Tools You Might Need

Not The Solution You Need? Search for Your Answer Here:

×

Login

Login or Sign up for access to all of our study tools and educational content!

Forgot password?
Register Now

×

Register

Sign up for access to all content on our site!

Or login if you already have an account

×

Reset password

If you have an active account we’ll send you an e-mail for password recovery

Or login if you have your password back