When titrating 50.0 mL of 0.10 M H2SO4 with 0.10 M NaOH, how many mL of NaOH will you have added to reach the 1st equivalence point?
I am confused... What H2A problem?
Chemistry-Dr.Bob222 - DrBob222, Tuesday, April 12, 2016 at 1:49am
Refer to the H2A problem above BUT you only titrate the first H on H2SO4. Post your work if you get stuck.
http://www.jiskha.com/display.cgi?id=1460431080
The problem you are referring to is a titration problem involving the reaction between 0.10 M H2SO4 (sulfuric acid) and 0.10 M NaOH (sodium hydroxide). The question asks for the volume of NaOH needed to reach the first equivalence point when 50.0 mL of the acid is titrated.
To solve this problem, we need to know the stoichiometry of the reaction. From the balanced equation:
H2SO4 + 2NaOH → Na2SO4 + 2H2O
We can see that it takes two moles of NaOH to neutralize one mole of H2SO4. Since the concentration of H2SO4 is 0.10 M, this means that 0.10 moles of H2SO4 are in every 1 L of the solution.
To find the number of moles of H2SO4 in the 50.0 mL solution, we need to calculate the moles per liter and then multiply by the volume:
moles of H2SO4 = (0.10 mol/L) × (50.0 mL / 1000 mL/L) = 0.005 mol
Since there is a 1:2 stoichiometric ratio between H2SO4 and NaOH, we know that we need twice as many moles of NaOH to reach the equivalence point. Therefore, we will need 0.010 mol of NaOH.
To find the volume of NaOH needed, we divide the number of moles by the concentration of NaOH:
volume of NaOH = (0.010 mol) / (0.10 mol/L) = 0.100 L or 100 mL
Therefore, you will need to add 100 mL of NaOH to reach the first equivalence point.