The volume of 10 ml of H2SO4 solution was diluted
250 ml with distilled water. This resulting solution, 25 ml were neutralized by 10 ml of 0.1 mol/L solution of NaOH. calculate the Molarity of sulfuric acid.
You need to be careful how you write questions; I think you meant that 10 mL H2SO4 was diluted TO 250 mL with DW. I'll assume that's what you meant.
2NaOH + H2SO4 ==> Na2SO4 + 2H2O
mols NaOH = M x L = 0.1 x 0.01 = 0.001.
mols H2SO4 in that 25 mL aliquot = 1/2 x 0.001 = 0.0005 (because you see 2 mol NaOH = 1 mol H2SO4).
Since that 25 mL aliquot was 1/10 of the 250 mL solution, the 0.0005 x 10 must be the amount inf the 250 mL flask. And that 0.005 mols H2SO4 must be what's in the 10 mL solution you started with. Now
M H2SO4 = mols H2SO4/L H2SO4 = 0.005/0.01 = ? Check my calculations. Easy to get mixed up when typing so many zeros. I like to work in millimols to avoid that problem; i.e.,
mmols NaOH = mL x M = 10 x 0.1 = 1
mmols H2SO4 = 1/2 that or 0.5 and that's the amount in the 25 mL aliquot. That makes 5 mmols (10x that) in the 10 mL.
Then M = mmols/mL = 5/10 = 0.5. I got the same answer. How about that.
To calculate the molarity of sulfuric acid (H2SO4), we can use the equation:
M1V1 = M2V2
Where:
M1 = initial molarity of H2SO4
V1 = initial volume of H2SO4
M2 = final molarity of NaOH
V2 = final volume of NaOH
Let's solve the problem step by step:
Step 1: Find the initial volume of H2SO4
The problem states that 10 ml of H2SO4 solution was initially diluted to a final volume of 250 ml. So, the initial volume of H2SO4 (V1) is 10 ml.
Step 2: Find the final volume of NaOH
The problem states that 25 ml of the resulting solution was neutralized by 10 ml of 0.1 mol/L NaOH. So, the final volume of NaOH (V2) is 10 ml.
Step 3: Find the final molarity of NaOH
The problem states that the molarity of NaOH (M2) is 0.1 mol/L.
Step 4: Calculate the initial molarity of H2SO4 (M1)
Using the equation M1V1 = M2V2, we can rearrange it to solve for M1:
M1 = (M2V2) / V1 = (0.1 mol/L * 10 ml) / 10 ml = 0.1 mol/L
Therefore, the molarity of the sulfuric acid solution is 0.1 mol/L.