Page 3 7/2/18 Name PART II. Please answer the following questions, SHOWING YOUR
ID: 704595 • Letter: P
Question
Page 3 7/2/18 Name PART II. Please answer the following questions, SHOWING YOUR WORK where required. NO WORK, NO CREDIT!! 1. What is the molarity of a solution made by dissolving 148 grams of NaCl in enough water to make 725 mL of solution? 2. What volume of 1.25 M BaCly do I need in order to prepare 350 mL of 0.75 M BaClh? 3. What mass of KOH do I need to prepare 1.5 liters of a 1.25 M solution of KOH? 4 In the following equation: NH4 Which is the conjugate acid? NH, ??. Which is the acid? 5. Write the squilibrium expression for question 4 above.Explanation / Answer
1. W = 148g
G.M.Wt of NaCl = 58.5g
volume of solution = 725ml
molarity = W*1000/G.M.Wt * volume of solution in ml
= 148*1000/58.5*725 = 3.5M
2. Before dilution After dilution
M1 = 1.25M M2 = 0.75M
V1 = V2 = 350ml
M1V1 = M2V2
V1 = M2V2/M1
= 0.75*350/1.25 = 210 ml
3. no of moles of KOH = molarity * volume in L
= 1.25*1.5 = 1.875 moles
mass of KOH = no of moles * gram molar mass
= 1.875*56 = 105g
mass of KOH = 105g
4.
NH3 + H2O -----------> NH4^+ + OH^-
Base Acid conjugate acid conjugate base
H2O is acid
NH4^+ is conjugate acid
5.
NH3 + H2O -----------> NH4^+ + OH^-
Kc = [NH4^+][OH^-]/[NH3]
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