Strong Acid-Weak Base Titrations. solving for concentration and pH?

Question

"if 86.5 ml of 0.002 M NaOH are required to reach the equilibrium point determine the concentration of the benzoic acid if you started the titration with 15 ml of benzoic acid. What will be the pH?"

I find this hard but for NaOH moles I got 0.00173 mol NaOH. And for Benzoic acid at 15ml i got 0.015 mol Benzoic acid.

I am stuck. I need help and directions.

Answer 1

Use this equation to find the second molarity.

molarity 1 x volulme 1 = molarity 2 x volume 2


The pH is the -log[H+]

Answer 2

For the titration of a weak acid with a strong base u can simply used this formula to figure out the concentration:
Co (Vo) = Ct ( Ve)
In other words, initial concentration of benzoic acid times the initial volume of bezoic acid = concentration of titrant times the volume of titratnt at equivalence pt. So
Co (.0150L)= (0.002M)(0.0865L)
Solve Co initial to be 0.01153 M benzoic acid
Then to solve for pH: you know that since this is an equivilance point situation, all the acid wil have reacted with your strong base so now all you have left is the product of that interation which is the conjugate base of benzoic acid (water is present also). So to calculate the pH u look at the dissociation of that conjugate base in water, get the OH- concentration and calculate pOH. then simply calculate
pH = 14 - pOH
The chemical amount of conjugate base that you want to look at is equal to 1.73x10^-4 mol since this is the amount produced at equivalence pt. So just check how much of OH- that yields. I hope u know how to solve for that . Just set up a chart. dont forget to convert the moles to molarities when solving for OH- concentration.

Answer 3

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