14. (a) How many milliliters of 0.155 M HCl are needed to neutralize completely

Question

14. (a) How many milliliters of 0.155 M HCl are needed to neutralize completely 45.0 mL of 0.101 M Ba(OH)2 solution?
ml


(b) How many milliliters of 3.50 M H2SO4 are needed to neutralize 75.0 g of NaOH?
mL


(c) If 54.8 mL of BaCl2 solution is needed to precipitate all the sulfate in a 554 mg sample of Na2SO4 (forming BaSO4), what is the molarity of the solution?
M


(d) If 27.5 mL of 0.125 M HCl solution is needed to neutralize a solution of Ca(OH)2, how many grams of Ca(OH)2 must be in the solution?

21. Using orbital diagrams, determine the number of unpaired electrons in each of the following atoms. Start with the noble-gas core designation in the first box and then enter electron arrows. Enter your answer using UP to indicate an upwards pointing arrow, DOWN to indicate a downwards pointing arrow, UP/DOWN to indicate two arrows(electrons) in the same orbital, and BLANK to indicate no arrows. Designate unpaired electrons as UP arrows and keep all unpaired electrons and empty orbitals in the rightmost boxes of a subshell. Fill in the last box in each atom with the number of unpaired electrons in the atom.

(a) Cs
[  ] s  d      p    
Number unpaired e-



(b) As
[  ] s  d      p    
Number unpaired e-



(c) Se
[  ] s  d      p    
Number unpaired e-



(d) Zn
[  ] s  d      p    
Number unpaired e-

Explanation / Answer

a. Ba(OH)2 + 2 HCl = BaCl2 + 2H2O

Thus 2 mol HCl is required to neutralise 1 mol of Ba(OH)2 solution. Thus 90 mL of 0.101M is required to netralise the Ba(OH)2 solution.

Now, use equation... V1 * S1 = V2 * S2

Thus, 90 * 0.101 = V2 * 0.155

or, V2 = 58.6 mL

Thus 58.6 mL of 0.155 M solution of HCl is required.

b. H2SO4 + 2NaOH = Na2SO4 + H2­O

Thus , 98 g of sulfuric acid is required to neutralise 80 g of NaOH.

So, to neutralise 75 g NaOH , 91.8 g of sulfuric acid is required.

Now 1000mL of 3.50 M H2SO4 = 98 * 3.5 = 343 g H2SO4

So, 1000*91.8/343 or 267.6 mL of 3.5 M H2SO4 solution will contain 91.8 g H2SO4 .

So, answer = 267.6 mL

c. BaCl2 + Na2SO4 = BaSO4 + 2 NaCl

Thus 208 g barium chloride reacts complrtely with 142 g sodium sulphate.

Thus, 0.554 g sodium sulphate will react completely with 0.554*208/142 or 0.8 g barium chloride.

So, 54.8 mL solution contain 0.8 g barium chloride.

So, 1000 mL of barium chloride solution = 0.8 * 1000 / 54.8 = 14.8 g barium chloride

So, molarity = 14.8/208 = 0.07 M

d. Ca(OH)2 + 2HCl = CaCl2 + 2 H2O

Thus 74 g Ca(OH)2 is needed to neutralise 73 g HCl.

Now, 27.5 mL of 0.125 M HCl = 73*27.5*0.125/1000 = 0.25 g HCl.

Amount of Ca(OH)2 present in the solution = 74 *0.25/73 = 0.254 g

Cs : [Xe] 6s1 , 1 unpaired electron in s orbital

As : [Ar] 4s2 4p3 , 3 unpaired electrons in p orbital

Se: [Ar] 4s2 4p4 , 2 unpaired electrons in p orbital

Zn: [Ar] 4s2 , No unpaired electron

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