20 cm3 of 1 M H\(_2\)SO\(_4\) is mixed 20 cm\(^3\) of 1 M NaOH. The resulting solution is?
acidic with a molarity of 0.50 M H2SO4
acidic with a molarity of 0.25 H2SO4
neutral
alkaline with a molarity of 0.25 M NaOH
alkaline with a molarity of 0.50 M NaOH
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New school is correct.
From the balanced equation
2NaOH+H2SO4===Na2SO4+2H2O
nNaOH=Cv
nNaOH=0.02×1=0.02moldm–³
nH2SO4=Cv
nH2SO4=0.02×1=0.02moldm–³
Since from the equation,
2moleNaOH=1MoleH2SO4
nNaOH=0.04moldm–³
nH2SO4=0.02moldm–³
0.04–0.02=0.02moldm–³
H2SO4 is the limiting reactant while NaOH is in excess
Total volume=(20+20)cm³
V=40cm³~~0.04dm³
n=CV
0.02=C×0.04
C=0.02/0.04
C=0.5M
yoo....cmon its neutral...they are playing with your heads....the acid and base have the same volume and conc .....strong acid vs strong base is neutral salt....pls some question in ume doesnt need solving
The reaction between sulfuric acid (H2SO4) and sodium hydroxide (NaOH) is a neutralization reaction, which is a type of acid-base reaction. The balanced chemical equation for this reaction is:
H2SO4+2NaOH→Na2SO4+2H2O
From the balanced equation, we can see that one mole of sulfuric acid reacts with two moles of sodium hydroxide to produce one mole of sodium sulfate and two moles of water.
Given that you have equal volumes (20 cm³) of 1 M solutions of both H2SO4 and NaOH, you have the same number of moles of H2SO4 and NaOH. However, because the stoichiometry of the reaction requires two moles of NaOH for every mole of H2SO4, all of the NaOH will react, but you will have some H2SO4 left over.
Therefore, the resulting solution will be acidic. To find the molarity of the remaining H2SO4, you would subtract the moles of H2SO4 that reacted (which is equal to the moles of NaOH) from the total moles of H2SO4. Then, divide by the total volume of the solution.
So, the correct answer is A. acidic with a molarity of 0.50 M H2SO4. This is because half of the H2SO4 did not react and is still present in the solution. The molarity is calculated as follows:
Molarity = Moles of solute / Volume of solution
= (0.5 * moles of H2SO4) / (20 cm³ + 20 cm³)
= 0.50 M
Moles of H₂SO₄:
Moles=Volume (in liters)×MolarityMoles=Volume (in liters)×Molarity
Moles=(20 cm3)×(1 mol/L)Moles=(20cm3)×(1mol/L)
Moles=20×10−3 molMoles=20×10−3mol
Moles of NaOH:
Moles=Volume (in liters)×MolarityMoles=Volume (in liters)×Molarity
Moles=(20 cm3)×(1 mol/L)Moles=(20cm3)×(1mol/L)
Moles=20×10−3 molMoles=20×10−3mol
Since the stoichiometry of the reaction between H₂SO₄ and NaOH is 1:2, we need twice as many moles of NaOH to react completely with the moles of H₂SO₄ present.
The remaining moles of NaOH after reaction:
Moles=(20×10−3 mol)−(20×10−3 mol)=0 molMoles=(20×10−3mol)−(20×10−3mol)=0mol
Now, the solution after reaction:
Since all the H₂SO₄ is neutralized, and there's no excess NaOH left, the resulting solution will be neutral.
Therefore, the correct answer is option C: neutral.
You guys don't really get, you are right to say that the solution should be neutral because it involves a strong acid and a strong base, but you should note and understand that according to the equation of the chemical reaction, 1 mole of H2SO4 requires 2 moles of NaOH for a complete neutralisation reaction, so if the number of moles is not complete, then the reaction can't be complete as to becoming neutral.
H2SO4 + 2NaOH ----> Na2SO4 + 2H2O
According to the question, the concentration and volume of the reactants are the same. 20cm³ and 1 molarity.
To calculate the mole, n = CV
H2SO4 = 1 × 0.02(V in dm³) = 0.02mol
NaOH = 1 × 0.02(V in dm³) = 0.02mol
This means 0.02 mol of H2SO4 reacts with 0.02mol of NaOH, this can never give us a neutral solution, because for this solution be neutral, we need the number of moles of NaOH to be twice that of H2SO4, so the number of moles off NaOH should've been 0.04mol. This would have resulted in a complete reaction because the number of moles needed by H2SO4 is complete.
But, we are having the same number of moles for both reactants,
H2SO4 + 2NaOH ----> Na2SO4 + 2H2O
number of moles - 0.02:0.02
reacting moles - 0.01:0.02
unreacted moles - 0.01:0
The limiting reactant is NaOH because it is not enough for the reaction, H2SO4 is the reactant I'm excess. 0.02 mol of NaOH can only react with 0.01 mol of H2SO4, to complete the reaction, we would need another 0.02 mol of NaOH to react with the remaining 0.01 mol of H2SO4, that's when we will have a complete neutralisation reaction. The total volume of the reaction is now 40cm³ (0.04dm³), i.e. 20cm³ H2SO4 + 20cm³ NaOH
To calculate the molarity (or concentration) of H2SO4, n = CV
C = n/V = 0.02/0.04 = 0.5M H2SO4
So, the correct answer is A) acidic with a molarity of 0.5M H2SO4
actually...it's ought to be B
from the equation of reaction
2NaOH + H2SO4 -------> Na2SO4 + 2H20
2mol of NaOH reacts with 1mol of H2S04
from the question...
the molarity of the reactants given is 1M and the volume = 4
20cm^3 = 0.02m^3
N = CV
n = 1 × 0.02
n = 0.02
thus,0.02mol of NaOH will react with 0.01mol of H2SO4,thereby leaving 0.01mol of H2SO4 unreacted in the solution mixture ,it will be acidic
to know the molarity
n = CV
where the volume in the reaction mixture = 20 + 20 = 40cm^3 ----> 0.04dm^3
0.01 = C × 0.04
C = 0.01/0.04
C = 0.25M
thus,the resultant solution will be acidic with a molarity of 0.25M ...
effect changes...thanks
using Ca x Va /Cb xVb = Na/Nb
1M x 20cm3 / 1x 20cm3 = 1/1
as you, they all cancel each other so, the answer for the question is neutral. according to my opinion and working out. thanks.
Answer should be B
First, write the reaction:
H2SO4 + 2NaOH =› Na2SO4 + 2H2O
Step 1: Calculate moles
Volume = 20 cm³ = 0.02 dm³
H₂SO₄:
1X 0.02= 0.02 mol
NaOH:
1X 0.02= 0.02 mol
Step 2: Consider stoichiometry
1 mole of H₂SO₄ reacts with 2 moles of NaOH
So:
0.02 mol H₂SO₄ needs 0.04 mol NaOH
But we only have 0.02 mol NaOH
NaOH is limiting, so not all acid is neutralized.
Step 3: Remaining acid
NaOH neutralizes half of the acid:
0.02 mol NaOH neutralizes 0.01 mol H₂SO₄
Remaining H₂SO₄:
0.02 – 0.01 = 0.01 mol
Step 4: Total volume :
20cm3 + 20cm3 = 40cm3 = 0.04dm3
Step 5: Final concentration:
0.01/0.04 = 0.25 M
Final Answer:
B. acidic with a molarity of 0.25 M H₂SO₄
For the acid
n = CV
n = 1×20/1000 = 0.02mol
For the base.
n = 1×20/1000 = 0.02mol.
These are the available moles.
But from the equation of reaction
Looking at the equation of reaction.
1 mole of H2SO4 requires 2 moles of NaOH
Therefore
0.02molesH2SO4 will require
0.02×2 = 0.04M NaOH
so NaOH is in excess by 0.02moles (i.e 0.04-0.02)
While H2SO4 was completely used up.
So, H2SO4 is the limiting reagent
Now let's know the concentration of the excess NaOH
n = CV
C = n/Total vol
n = 0.02moles(concentration of NaOH at the end of the reaction
V = 20+20 = 40cm³ = 0.04dm³
C = 0.02/0.04 = 0.5M
So it's the excess NaOH that furnishes the chemical entity that furnishes the pH of the resulting solution.
NaOH -----> OH- + Na+
POH -Log(OH-)
POH = -Log(0.5) = 0.3010
PH + POH = 14
PH = 14-0.3010 = 13.699
So the solution is alkaline with a molarity of 0.5M NaOH
E 
Myschool is right
H2SO4 + 2NaOH-->Na2SO4 + 2H2O
1 : 2 -->
20 20
10 20
showing that the 20cm³ of NaOH only need 10cm³ of the acid.. which is half of the given acid and half of given concentrated
1M of NaOH will only need 0.5M acid
0.5M of acid in excess
The ans they gave is correct. It's acidic with 0.5M H2SO4. Because sulphuric acid is in excess by 0.01 mol
Weak acid + Strong base
→ Basic salt + Water
Strong acid + Weak base
→Acidic salt + Water
Strong Acid + Strong Base
→Neutral salt + Water
Since, hydrochloric acid is a strong acid and sodium hydroxide is a strong base, therefore, their reaction produces neutral
Correct option is C)
Concept Applied: A neutralization reaction is when an acid and a base react to form water and salt and involves the combination of hydrogen ions and hydroxyl ions to generate water. The neutralization of a strong acid and strong base has a pH equal to 7.
Correct Option: C in
Explanation for correct option: The complete neutralization of a strong acid (H2SO4) and strong base ( NaOH ) has a pH equal to 7.
Explanation for incorrect options: (H2SO4) and NaOH both have the same concentration so they neutralized each other and make ph equal to 7 so rest all other options are wrong.
