4) What mass of CuSO 4 .5H 2 O (in grams) do you need to make 250.0 mL of a 1.50
ID: 714929 • Letter: 4
Question
4) What mass of CuSO4.5H2O (in grams) do you need to make 250.0 mL of a 1.50 M CuSO4 solution?
5) Calculate the mass of Sodium Phosphate trihydrate, Na3PO4 .3H2O, that is needed to make 100.00 ml of 0.250M Na3PO4 solution.
6) Calculate the concentration of sodium ion in the solution made in question #5.
7) To what volume (in mL) should you dilute 100.0 mL of a 5.00 M CaCl2 solution to obtain a 0.750 M CaCl2 solution?
8) What volume of a 6.00 M NaNO3 solution should you use to make 0.525 L of a 1.20 M NaNO3 solution?
9) 45.00 ml of water is added to a sample of 5.00 ml HCl stock solution. The concentration of HCl in diluted solution was determined to be 0.365 M. Calculate the concentration of HCl in stock solution.
10) Calculate the volume of 0.200 M HCl (acid) is used to neutralize 3.00 grams of ml of NaOH (base). Write the balanced chemical reaction first.
11) 13.5 ml of 0.125M NaOH (base) is needed to neutralize 25.00 ml of sulfuric acid (H2SO4). Calculate the Molarity of Sulfuric acid solution.
12) 10.0 ml of 0.200 M sodium Hydroxide (NaOH) is added to 25.00 ml of 0.100M hydrochloric acid solution (HCl). Which one of the reactants, (NaOH), or (HCl) will be left over after the neutralization is completed. (show your calculation)
13) Ionic compounds dissociate into ions completely when dissolved in water. Determine the concentration (molarity) of the indicated ions in the following solution.
a) Acetate ion (C2H3O2)) in 2.00 M Sodium acetate (NaC2H3O2)
b) Ammonium ion (NH4+) in 2.00M ammonium sulfate ((NH4)2SO4
c) Phosphate ion (PO43-) in 2.00M Ca3(PO4)2
d) Calcium ion in in 2.00M Ca3(PO4)2
Explanation / Answer
Solution :-
4) What mass of CuSO4.5H2O (in grams) do you need to make 250.0 mL of a 1.50 M CuSO4 solution?
Solution :-
Using the molarity and volume we can find the moles of CuSO4.5H2O needed
Moles = molarity x volume in liter
= 1.50 mol per L * 0.250L
= 0.375 mol CuSO4.5H2O
Molar mass of CuSO4.5H2O= 249.685 g/mol
Mass = moles x molar mass
Mass of CuSO4.5H2O = 0.375 mol * 249.685 g per mol
= 93.6 g
Therefore the mass of CuSO4.5H2O needed to make 1.5 M solution of CuSO4 of total volume 250 ml is 93.6 g CuSO4.5H2O
5) Calculate the mass of Sodium Phosphate trihydrate, Na3PO4 .3H2O, that is needed to make 100.00 ml of 0.250M Na3PO4 solution.
Solution :-
Moles = molarity x volume
= 0.250 mol per L * 0.100 L
= 0.0250 mol
Molar mass of Na3PO4.3H2O = 217.9865 g per mol
Mass of Na3PO4.3H2O = 0.0250 mol *217.9865 g per mol
= 5.45 g
Therefore mass of Na3PO4.3H2O needed is 5.45 g
6) Calculate the concentration of sodium ion in the solution made in question #5.
Na3PO4 dissociates as follows
Na3PO4 ----> 3Na^+ + PO4^3-
Therefore mole ratio of the Na3PO4 to Na^+ is 1 : 3
Therefore concentration of the Na^+ ion is triple than concentration of the Na3PO4
[Na^+] = 3*[Na3PO4]
= 3*0.250 M
= 0.750 M
Therefore sodium ion concentration is 0.750 M
7) To what volume (in mL) should you dilute 100.0 mL of a 5.00 M CaCl2 solution to obtain a 0.750 M CaCl2 solution?
Solution :-
Initial molarity M1 = 5.00 M
Initial volume = 100.0 ml
Final molarity M2=0.750 M
Final volume V2= ?
M1V1=M2V2
V2=M1V1/M2
= 5.00 M * 100.0 ml / 0.750 M
= 667 ml
Therefore we need to dilute the solution to the final volume of 667 ml
8) What volume of a 6.00 M NaNO3 solution should you use to make 0.525 L of a 1.20 M NaNO3 solution?
Solution :-
M1 = 6.00 M
V1= ?
V2=0.525 L
M2 = 1.20 M
M1V1=M2V2
V1= M2V2/M1
= 1.20 M * 0.525 L / 6.00 M
= 0.105 L
Therefore we need 0.105 L of 6.00 M NaNO3 solution.
9) 45.00 ml of water is added to a sample of 5.00 ml HCl stock solution. The concentration of HCl in diluted solution was determined to be 0.365 M. Calculate the concentration of HCl in stock solution.
Solution :-
Initial volume V1 = 5.00 ml
Initial molarity M1=?
Final molarity M2= 0.365 M
Final volume V2 = 45.00 ml + 5.0 ml = 50 ml
M1V1=M2V2
M1 = M2V2/V1
= 0.365 M*50.0 ml / 5.00 ml
= 3.65 M
Therefore molarity of the stock solution of HCl is 3.65 M
10) Calculate the volume of 0.200 M HCl (acid) is used to neutralize 3.00 grams of ml of NaOH (base). Write the balanced chemical reaction first.
Solution :-
Balanced reaction equation
HCl + NaOH ---> NaCl + H2O
Moles of NaOH = mass / molar mass
= 3.00 g / 40.0 g per mol
= 0.075 mol
Since mole ratio of the HCl and NaOH is 1:1 therefore moles of HCl needed are same
So we need 0.075 mol HCl
Molarity = moles / volume
Volume of HCl = moles / molarity
= 0.075 mol / 0.200 mol per L
=0.375 L
Therefore it needs 0.375 L HCl
11) 13.5 ml of 0.125M NaOH (base) is needed to neutralize 25.00 ml of sulfuric acid (H2SO4). Calculate the Molarity of Sulfuric acid solution.
Solution :-
H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
Moles of NaOH = molarity x volume
= 0.125 mol per L * 0.0135 L
= 0.001688 mol NaOH
mole ratio of the NaOH and H2SO4 is 2 : 1
Lets calculate the moles of H2SO4 using the moles of NaOH
0.001688 mol NaOH * 1 mol H2SO4 / 2 mol NaOH = 0.000844 mol H2SO4
Molarity = moles /volume
Molarity of H2SO4 = 0.000844 mol / 0.025 L
= 0.0338 M
12) 10.0 ml of 0.200 M sodium Hydroxide (NaOH) is added to 25.00 ml of 0.100M hydrochloric acid solution (HCl). Which one of the reactants, (NaOH), or (HCl) will be left over after the neutralization is completed. (show your calculation)
Solution :-
NaOH + HCl --- > NaCl + H2O
Moles of NaOH = 0.200 mol per L * 0.0100 L = 0.0020 mol NaOH
Moles of HCl = 0.100 mol per L * 0.025 L = 0.0025 mol HCl
Mole ratio of the HCl and NaOH is 1 :1
Therefore since moles of NaOH are less than moles of HCl hence the NaOH is limiting reactant and HCl is excess reactant therefore at the end of reaction HCl will be left over.
13) Ionic compounds dissociate into ions completely when dissolved in water. Determine the concentration (molarity) of the indicated ions in the following solution.
a) Acetate ion (C2H3O2)) in 2.00 M Sodium acetate (NaC2H3O2)
NaC2H3O2 ---> Na^+ + C2H3O2^-
1 mol sodium acetate gives 1 mole acetate ion
Therefore molarity of the sodium acetate and acetate ion will be same
2.00 M NaC2H3O2 * 1 acetate ion / 1 NaC2H3O2 = 2.00 M C2H3O2^-
b) Ammonium ion (NH4+) in 2.00M ammonium sulfate ((NH4)2SO4
1 mol ammonium sulfate gives 2 mol ammonium ion
2.00 M (NH4)2SO4 * 2 NH4^+ / 1 (NH4)2SO4 = 4.00 M NH4^+
Therefore concentration of ammonium ion is 4.00M
c) Phosphate ion (PO43-) in 2.00M Ca3(PO4)2
1 mol Ca3(PO4)2 = 2 mol PO4^3-
2.00 M Ca3(PO4)2 * 2 PO4^3- / 1 Ca3(PO4)2 = 4.00 M PO4^3-
Therefore concentration of the phosphate ion is 4.00 M
d) Calcium ion in in 2.00M Ca3(PO4)2
1 mol Ca3(PO4)2 = 3 mol Ca^2+
2.00 M Ca3(PO4)2 * 3 Ca^2+ / 1 Ca3(PO4)2 = 6.00 M Ca^2+
Therefore concentration of the calcium ion us 6.00 M
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