
Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community for help and clear up your study doubts
Discover the best universities in your country according to Docsity users
Free resources
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Material Type: Quiz; Class: Intro to Chemical Principles; Subject: Chemistry; University: City College of San Francisco; Term: Fall 2008;
Typology: Quizzes
1 / 1
This page cannot be seen from the preview
Don't miss anything!
Chemistry 40 Name______________________________ Quiz 4, Fall 08 1.) Complete the following table. (3points) Acid Base Conjugate Acid Conjugate Base H 2 O NH 3 NH 4 +^ HO - HCl H 2 O H 3 O +^ Cl- H 2 SO 4 NaOH H 2 O HSO 4 -
2.) What is the molarity of a 200ml solution containing 4.2 g of NaCl? (3 points) Molar Mass of NaCl = 58.443g moles of NaCl = 4.2g / 58.433g = .072 moles Molarity (M) = moles (n)/ volume (L) = .072 / 200 x 10 -3^ = .36M 3.) What is the concentration of potassium ions in a 1.4M solution of K 2 SO 4? (2 points) K 2 SO 4 = 2K+^ + SO 4 2- 1.4M x 2 = 2.8M 5.) What is the pH of a 10 - 3^ M solution of HCl? (2 points) pH = 3 6.) What is the [H +] of a solution given pOH = 3.28 (2 points) pH + pOH = 14 pH = 14 – 3.28 = 10. [H +] = 1.9 x 10- 7.) Calculate the pH of a 40 ml solution containing 1.00 g of nitric acid. (3 points) Nitric acid = HNO 3 HNO 3 = H +^ + NO 3 - Molar Mass of HNO 3 = 63.012g moles of HNO 3 = 1.00g / 63.012g = .0159 moles Molarity (M) = moles (n)/ volume (L) = .0159 / 40 x 10 -3^ = .398M
8.) How many ml of a 0.2 M NaOH solution are required to bring the pH of 20 ml of a 0.4 M HCl solutionto 7.0? (5 points)
At pH = 7 [H +] = [OH -^ ] Molarity (M) = moles (n)/ volume (L) moles of HCl = 0.4M x 20x10 -3^ = .008 = moles of NaOH needed Volume (L) = .008 moles of NaoH / 0.2M = .04 L or 40 ml