Can someone help with my chemistry problems homework? I am struggling to get the
ID: 1049731 • Letter: C
Question
Can someone help with my chemistry problems homework? I am struggling to get the answers to these and need a little assistance please! I would greatly appreciate any help anyone can offer to me for these problems!
14. (5pnts) What is the molarity of potassium ion in the following solutions: a. 0.20 M potassium chloride b. 0.15 M potassium chromate c. 0.080 M potassium phosphate 15. (5pnts) How many mL of 6.0 M nitric acid are needed to prepare 100.0 mL of 0.500 M nitric acid? 16. (5pnts) Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1.049 g/mL. Calculate the molarity of a solution of acetic acid made by dissolving 20.00 mL of glacial acetic acid in enough water to make 250.0 mL of solution. 17. (5pnts) How many milliliters of 0.128 M hydrochloric acid is needed to neutralize 2.87 g of solid magnesium hydroxide? 18. (5pnts) If 25.8 mL of a silver nitrate solution is needed to precipitate all the chloride ions in a 785- mg sample of KCI (forming silver chloride), what is the molarity of the AgNO3 solution?Explanation / Answer
Ans. 14.
a. KCl ---------> K+ + Cl-
1 mole of KCl produces 1 mole of K+ ions. Thus, molarity of potassium is equal to that of KCl.
So, [K+] = 0.2 M
b. K2CrO4 -------------> 2K+ + CrO42-
1 mole of K2CrO4 produces 2 mole of K+ ions. Thus, molarity of potassium is equal to twice the molarity of K2CrO4.
So, [K+] = 2 x 0.15 M = 0.3 M
c. KH2PO4
1 mole of KH2PO4 produces 1 mole of K+ ions. Thus, molarity of potassium is equal to that of KH2PO4.
So, [K+] = 0.080 M
Ans. 15. Using: M1V1 = M2V2 --- equation 1
Where, M1= molarity of solution 1, V1= volume of solution 1 (say, 6.0 M nitric acid)
M2= molarity of solution 2, V2= volume of solution 2 (say, 0.5 M nitric acid)
Or, 6.0 M x V1 = 0.5 M x 100 mL
Or, V1 = (50 M L) / (6.0 M) = 8.33 mL
Thus, required volume of 6.0 M nitric acid = 8.3 mL
Ans. 16. Given, density of glacial acetic acid = 1.049 g/ mL
Mass of 20 mL acetic acid = density x volume
= 1.049 g/ mL x 20 mL = 20.98 g
Molecular mass of acetic acid = 60.052 g/ mol
Number of moles of acetic acid = mass/ molecular mass
= 20.98 g / (60.052 g/ mol)
= 0.34936 moles
Molarity of acetic acid solution = number of moles of acetic acid / volume in L
= 0.34936 moles / 0.250 L
= 1.397 moles/ L = 1.397 M
Ans. 17. Number of moles of Mg(OH)2 in 2.87-gram sample =
Mass / molecular mass
= 2.87 g/ (58.319 g/mol) = 0.049212 moles
Since, 1 mole of Mg(OH)2 donates 2 OH- ions, 1 mole Mg(OH)2 requires 2 moles of HCl to get neutralized.
Thus, number of moles of Mg(OH)2 = 2 x number of moles of HCl
Thus, required number of moles of HCl = 2 x 0.049212 moles = 0.0984 moles.
Now, using,
Number of moles = molarity x volume in L
Or, 0.0984 moles = 0.128 M x L = (0.128 moles/ L) x V
Or, V = (0.0984 moles) / (0.128 moles/ L) = 0.76893 L = 768.93 mL
Ans. 18. Moles of KCl = mass / molecular mass
= 0.785 g / (74.548 g/ mol) [1 g = 1000 mg]
= 0.01053 moles
Now, AgNO3(aq) + KCl(aq) -----------> AgCl(s) + KNO3(aq)
1 mol of KCl reacts with 1 mol AgNO3 to precipitates 1 mol of AgCl.
Thus, moles of AgNO3 required to all chlorides in KCl sample to precipitate = 0.01053 moles
Using, Number of moles = molarity x volume in L
Or, 0.01053 moles = M x 0.0258 L [1 L = 1000 mL]
Or, M = 0.4081445 moles/ L = 0.41 M