Chemistry problem about solutions and pH.

Question

"Calculate how many cm^3 of a solution of sulfuric acid 80% (density = 1.727 g/mL) must be diluted with water in order to prepare a 0.2 L 0.4 M solution. If we add to this final solution of sulfuric acid 800 cm^3 of water and 6g of NaOH: What's the pH?".

 

I hope someone will help me because I don't understand how to solve it!

Answer 1

There are two questions need to solve. 1. How many ml/cm³ 80% H₂SO₄ we need. 2. What is the PH.

 

1.To dilute the H₂SO₄ from 80% solution, only need to add water, so the number/mass of the H₂SO₄ molecule maintain the same. 

2.The H₂SO₄ we use to make reaction with NaOH is the 0.2L 0.4M solution. So we need to know how much H₂SO₄ in the solution. For NaOH, it use 6g (which is mass), so we need to find out how many g (which is mass) of H₂SO₄ there is.

 

For you can see now, the mass of H_²SO₄ is the key for both questions.

 

1. 0.2L 0.4M solution

Molar mass of H=1 g/mol,S=32 g/mol,O=16 g/mol

Molar mass of H₂SO₄=1*2+32+16*4=98 g/mol

 

M is Molarity, Molarity =Mols/Liters

the mols number of H₂SO₄=Molarity*Liters=0.4*0.2=0.08 mol

 

Mass=Mols number* Molar mass

Mass of H₂SO₄= the mols number of H₂SO₄* Molar mass of H₂SO₄

=98 g/mol* 0.08 mol=7.84 g

 

Mass of H₂SO₄/ Mass of solution=80%

Mass of solution=7.84 g/0.8=9.8 g

Volume of solution=Mass / Density=9.8 g/1.727 g/ml=5.67 ml

 

2. H₂SO₄+2NaOH=Na₂SO₄+2H₂O

 

Molar mass of H₂SO₄=98 g/mol

Molar mass of NaOH=23+16+1=40 g/mol

Molar mass of H₂SO₄/ Molar mass of 2NaOH= Mass of H₂SO₄/ Mass of NaOH

98/2*40=Mass of H₂SO₄/6 g

Mass of H₂SO₄ need in this reaction=7.35 g

Total Mass of H₂SO₄ in the solution is 7.84 g

It means after the reaction H₂SO₄ left by, 7.84-7.35=0.49 g

the mols number of left H₂SO₄=Mass/Molar mass of H₂SO₄=0.49 g/ 98 g/mol=0.005 mol

Each H₂SO₄ has two H⁺

so the total mols number of H⁺ is 0.005 mol*2=0.01 mol

 

Because we added 800 ml water, and the H₂SO₄ solution was 200 ml, so there is o.o1 mol H⁺ in 1000 ml solution

Here comes Molarity again.

Molarity =Mols/Liters

0.01 mol/1 L=0.01 M

 

PH=-log [H⁺]=-log (0.01)= -log (0.1)²=-(-2)=2

 

Answer 2

First calculate the molarity of the H2SO4 solution:
Take 1.0L of the solution = 1000 mL
mass = Volume * density
Mass = 1000mL * 1.727g/mL
Mass = 1727g
Mass of H2SO4 = 80/100*1727 = 1,381.6 g H2SO4
Molar mass H2SO4 = 50g/mol
mol H2SO4 in 1,381.6g = 1381.6/50 = 27.6 mol H2SO4 in 1.0L solution
Molarity = 27.6M 
To make 0.2L = 200mL of 0.4M solution: 
M1V1 = M2V2
27.6*V1 = 0.4*200
V1 = 0.4*200/27.6
V1 = 2.9 mL of the concentrated acid required.

Now dilute this solution to 200+800 = 1000mL
Molarity of solution
M1V1 = M2V2
M1*1000 = 0.4*200
M1 = 0.4*200/1000
M1 = 0.08M H2SO4
Because the volume is 1.0L - you have 0.08 mol H2SO4 in solution
Equation:
2NaOH + H2SO4 → Na2SO4 + 2H2O
1 mol H2SO4 will react with 2 mol NaOH
You have 0.08 mol H2SO4 - this will react with 0.08*2 = 0.16 mol NaOH
Molar mass NaOH = 40g/mol
mol NaOH in 6.0g = 6/40 = 0.15 mol
The NaOH is insufficient to react with all the H2SO4
0.15 mol NaOH will react with 0.075 mol H2SO4
You have 0.08 - 0.075 = 0.005 mol H2SO4 dissolved in 1.0L solution .
If we assume that the H2SO4 dissociates completely :
[H+] = 2*[H2SO4]
[H+] = 2*0.005
[H+] = 0.01M
pH = -log [H+]
pH = -log 0.01
pH = 2.00