Can someone help me with this pleas? A chemist needs to determine the concentration of a sulfuric acid solution by titration with a standard sodium hydroxide
solution. He has a 0.1598 M standard sodium hydroxide solution. He takes a 25.00 mL sample of the original acid solution
and dilutes it to 250.0 mL. Then, he takes a 10.00 mL sample of the dilute acid solution and titrates it with the standard
solution. The endpoint was reached after the addition of 11.55 mL of the standard solution. What is the concentration of the
original sulfuric acid solution?

For this problem, we have to calculate for the concentration of sulfuric acid using a standard 0.1820 M NaOH

Recall that at the equivalence point of a titration:

moles acid=moles base

Also, recall that moles = molarity × volume. 

This means:

MVacid=MVbase

We can use this equation to calculate the molarity of H2SO4. We’re given the following values:

Macid = ??                 Mbase = 0.182 M NaOH

Vacid = 25.0 mL         Vbase = 13.8 mL

Solving for Macid:

 Macid10.0 mL=0.182 M NaOH13.89 mLMacid = (0.182 M)(13.89 mL)10 mL

Macid= 0.2528 M HCl

Then use dilution equation to to find the molarity of the original sample (25 mL). Use the diluted amount of 250 mL

(MV)initial = (MV)dilutedMinitial(25 mL)=(250 mL)(0.2528 M)Minitial=(250 mL)(0.2528 M)25 mL

Minitial = 2.523 M

Therefore, the concentration of the HCl solution is 0.07726 M.

the answer is 0.8m.

by: nicholas