Question

Home assignment: Chapter 4: Aqueous Reactions and Solution Stoichiometry 17) An aliquot (28.7 ml.) of a KOH solution required 31.3 mL of 0.118 M HCl for neutralization. What mass (g) of KOH was in the original sample? A) 1.64 B) 7.28 C) 0.173 D) 0.207 E) 0.414 18) Which of the following would require the largest volume of 0.100 M sodium hydroxide solution for neutralization? A) 10,0 mL of 0.0500 M phosphoric acid B) 20,0 mL of 0.0500 M nitric acid C) 5.0 mL of 0.0100 M sulfuric acid D) 15,0 mL of 0.0500 M hydrobromic acid E) 10.0 mL of 0.0500 M perchloric acid

please answer all questions and show work?

Answer 1

Answer:-

These questions are answered by using simple concept of stoichiometry and the simple formula of moles and mass.

The answer is given in the image,

Annuer:- 7) HA+KOH ke +H2O Mdesa kok = Molesos HCl = 0.118 x 31.3 looo 53 6934X 16-3mol. Males malso KOH= 39+16 +1 =56g/mol massa kol= molés x molar mass 23.6934 X10-3x56 = 0.2079. D) 0.207 is correct 18) A).10.0mL of 0.0500 Mphosphoricacid because it will most itt moles.