A 21.18 mL of 0.250 M NaOH is titrated with a H2SO4 solution. The initial
volume of H2SO4 was 13.28 and the final volume of H2SO4 was 28.29 mL when
the solution turned very slightly pink. What is the concentration of H2SO4?
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Ask questions about the solution. Reposting it is not likely to get you much help. I assume you have had the titration equation.
To determine the concentration of H2SO4, you can use the concept of neutralization and stoichiometry. Here's how you can calculate it step by step:
Step 1: Identify the balanced chemical equation for the reaction between NaOH and H2SO4. The balanced equation is:
2 NaOH + H2SO4 -> Na2SO4 + 2 H2O
This equation tells us that 2 moles of NaOH react with 1 mole of H2SO4.
Step 2: Determine the moles of NaOH used in the titration. To do this, multiply the volume of NaOH solution used (21.18 mL) by its concentration (0.250 M) and convert it to moles:
Moles of NaOH = (Volume of NaOH solution in L) x (Concentration of NaOH in M)
= (21.18 mL / 1000 mL/L) x (0.250 mol/L)
= 0.005295 mol NaOH
Step 3: Apply the stoichiometry of the balanced equation to find out the moles of H2SO4 that reacted with NaOH. Since the stoichiometry ratio is 1:1 between H2SO4 and NaOH, the moles of H2SO4 will be the same as the moles of NaOH:
Moles of H2SO4 = 0.005295 mol H2SO4
Step 4: Calculate the concentration of H2SO4. The volume of H2SO4 solution used in the titration was determined by subtracting the initial volume (13.28 mL) from the final volume (28.29 mL):
Volume of H2SO4 used = Final volume - Initial volume
= 28.29 mL - 13.28 mL
= 15.01 mL = 0.01501 L
Now, divide the moles of H2SO4 by the volume of H2SO4 used to obtain the concentration:
Concentration of H2SO4 = Moles of H2SO4 / Volume of H2SO4 used
= 0.005295 mol / 0.01501 L
= 0.352 M
So, the concentration of H2SO4 is approximately 0.352 M.