100ml of 0.5M H2SO4 solution is neutralized by
50ml of 0.1M NaOH and x ml of 0.1M Ca(OH)2
The value of x is
(1) 200 ml
(2) 370 ml
(3) 475 ml
(4) 405 ml
I know answer is 3 I want the complete solution
Answers
Both Na2CO3 and NaOH are involved in the reaction:
2NaOH + H2SO4 -------------> Na2SO4 + 2H2O
Ca(OH)2 + H2SO4 -------------> CaSO4 + 2H2O
100 mL of 0.5 M H2SO4 is used.
0.1 L of 0.5 M H2SO4 is used.
molarity = (moles of H2SO4) / (liters of solution)
0.5 = (moles of H2SO4) / 0.1
moles of H2SO4 = 0.5 x 0.1 => 0.05 moles of H2SO4.
As 50 mL of 0.1 M NaOH is utilized completely.
0.05 liters of 0.1 M NaOH
0.1 = (moles of NaOH)/0.05
moles of NaOH = 0.05 x 0.1 => 0.005 moles of NaOH
First, let us try to find out how much of the H2SO4 is neutralized by 0.005 moles of NaOH.
The reaction is : 2NaOH + H2SO4 -------------> Na2SO4 + 2H2O
Thus, 2 parts of NaOH react with 1 part of H2SO4.
Hence, 2 parts of NaOH is 0.005 moles of NaOH
1 part = 0.005/2 => 0.0025 moles
Hence, 1 part of H2SO4 is 1 x 0.0025 => 0.0025 moles of H2SO4.
Hence, only 0.0025 moles of H2SO4 is neutralized.
Hence, 0.05 - 0.0025 => 0.0475 moles of H2SO4 remains in solution.
Now, we have to find the number of moles of Ca(OH)2 that neutralize 0.0475 moles of H2SO4.
The equation is : Ca(OH)2 + H2SO4-----------> CaSO4 +2H2O
Hence, 1 part of Ca(OH)2 reacts with 1 part of H2SO4.
Hence, 1 part of H2SO4 is 0.0475 moles.
So, 1 part of Ca(OH)2 is 0.0475 moles.
So, 0.0475 moles of Ca(OH)2 should be present in the solution.
Volume = x ml => x/1000 L
Moles of solute = 0.0475 moles
Molarity = 0.1 M
Molarity = (moles of Ca(OH)2) / (liters of solution)
0.1 = 0.0475 / (x/1000)
0.1 = 0.0475 * 1000/x
0.1 = 47.5 /x
x=47.5/0.1
x= 47.5 * 10
x = 475 mL
Hence, 475 mL of 0.1 M Ca(OH)2 will be needed.
So the correct option is 3) 475 mL
Answer:
475ml
Explanation:
Both Na2CO3 and NaOH are involved in the reaction:
2NaOH + H2SO4 -------------> Na2SO4 + 2H2O
Ca(OH)2 + H2SO4 -------------> CaSO4 + 2H2O
100 mL of 0.5 M H2SO4 is used.
0.1 L of 0.5 M H2SO4 is used.
molarity = (moles of H2SO4) / (liters of solution)
0.5 = (moles of H2SO4) / 0.1
moles of H2SO4 = 0.5 x 0.1 => 0.05 moles of H2SO4.
As 50 mL of 0.1 M NaOH is utilized completely.
0.05 liters of 0.1 M NaOH
0.1 = (moles of NaOH)/0.05
moles of NaOH = 0.05 x 0.1 => 0.005 moles of NaOH
First, let us try to find out how much of the H2SO4 is neutralized by 0.005 moles of NaOH.
The reaction is : 2NaOH + H2SO4 -------------> Na2SO4 + 2H2O
Thus, 2 parts of NaOH react with 1 part of H2SO4.
Hence, 2 parts of NaOH is 0.005 moles of NaOH
1 part = 0.005/2 => 0.0025 moles
Hence, 1 part of H2SO4 is 1 x 0.0025 => 0.0025 moles of H2SO4.
Hence, only 0.0025 moles of H2SO4 is neutralized.
Hence, 0.05 - 0.0025 => 0.0475 moles of H2SO4 remains in solution.
Now, we have to find the number of moles of Ca(OH)2 that neutralize 0.0475 moles of H2SO4.
The equation is : Ca(OH)2 + H2SO4-----------> CaSO4 +2H2O
Hence, 1 part of Ca(OH)2 reacts with 1 part of H2SO4.
Hence, 1 part of H2SO4 is 0.0475 moles.
So, 1 part of Ca(OH)2 is 0.0475 moles.
So, 0.0475 moles of Ca(OH)2 should be present in the solution.
Volume = x ml => x/1000 L
Moles of solute = 0.0475 moles
Molarity = 0.1 M
Molarity = (moles of Ca(OH)2) / (liters of solution)
0.1 = 0.0475 / (x/1000)
0.1 = 0.0475 * 1000/x
0.1 = 47.5 /x
x=47.5/0.1
x= 47.5 * 10
x = 475 mL
Hence, 475 mL of 0.1 M Ca(OH)2 will be needed.
So the correct option is 3) 475 mL