Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 0.080 M NaOH.
A) benzoic acid (C6H5COOH) Ka= 6.3*10^-5 for it
thanx
To calculate the pH at the equivalence point in titrating benzoic acid (C6H5COOH) with 0.080 M NaOH, we need to determine the moles of benzoic acid and NaOH at the equivalence point.
First, let's determine the moles of benzoic acid (C6H5COOH) in the 0.100 M solution. The molar mass of benzoic acid is 122.12 g/mol. Therefore, the moles of benzoic acid can be calculated as follows:
Moles of benzoic acid = (0.100 M) x (volume of solution in liters)
Next, we need to determine the moles of NaOH (sodium hydroxide) required to reach the equivalence point. The balanced chemical equation for the reaction between benzoic acid and NaOH is:
C6H5COOH + NaOH → C6H5COONa + H2O
According to the balanced equation, one mole of benzoic acid reacts with one mole of NaOH. Therefore, the moles of NaOH at the equivalence point will be equal to the moles of benzoic acid.
Once we know the moles of NaOH, we can calculate the concentration of NaOH at the equivalence point:
Concentration of NaOH = Moles of NaOH / (volume of solution in liters)
Now that we have the concentration of NaOH at the equivalence point, we can calculate the pOH using the concentration of hydroxide ions (OH-) in the solution:
pOH = -log(OH- concentration)
Finally, we can calculate the pH at the equivalence point using the pOH:
pH = 14 - pOH
Remember to perform the calculations using appropriate significant figures and units.
ph = 5.57
Didn't I show you how to do the other two of the three you posted last night? I think so. Now I've worked all three for you. This one is worked exactly like part b of your three part post.
If we call this HB (for benzoic acid), then
..........B^- + HOH ==> HB + OH^- initial.0.0444...........0....0
change....-x.............x.....x
equil..0.0444-x...........x....x
Kb = Kw/Ka = (HB)(OH^-)/(B^-)
Substitute from the ICE chart, solve for OH^- and convert to pH.