A 50.0 mL sample of 0.42 M benzoic acid, C6H5COOH, a weak monoprotic acid, is titrated with 0.40 M NaOH. Calculate the pH at the equivalence point. Ka of C6H5COOH = 6.5 10-5.
I know I have to set up an ICE chart, but I keep getting the wrong answer
At the equivalence point you have sodium benzoate, the salt. Hydrolyze the salt and do the ICE.
C6H5COONa + HOH ==> C6H5COOH + OH^-
Kb for the benzoate = (Kw/Ka)
(Kw/Ka) = (C6H5COOH)(OH^-)/(C6H5COONa)
(OH^-)= x = (C655COOH)
Plug in (C6H5COONa) from the titration data and solve for x which = (OH^-), then convert to pH.
Do you mind substiuting in numbers? I am still kinda confused...sorry.
nvm got it. thanks!
To calculate the pH at the equivalence point, you need to understand the process of the titration and the behavior of the weak acid and the strong base.
First, let's set up an ICE (Initial, Change, Equilibrium) table to keep track of the concentrations of the reactants and products during the titration.
Initial:
- Benzoic acid (C6H5COOH): 0.42 M × 50.0 mL = 0.021 mol
- Water (H2O): 50.0 mL = 0.050 L
Change:
- Benzoic acid: -x (as it reacts with NaOH)
- NaOH: -x (as it reacts with benzoic acid)
- Water: +x (as NaOH is added)
Equilibrium:
- Benzoic acid: 0.021 - x
- NaOH: x
- Water: 0.050 + x
From the balanced equation of the reaction between benzoic acid and NaOH, you know that 1 mol of benzoic acid reacts with 1 mol of NaOH. So, the final concentrations of benzoic acid and NaOH at the equivalence point will be equal.
Now, you can set up the equation that relates the concentration of the benzoic acid to the concentration of hydroxide ions (OH-) at the equivalence point:
Ka = [C6H5COOH] / [OH-]
Plug in the values:
6.5 × 10^-5 = 0.021 - x / x
Solve for x:
x = (6.5 × 10^-5) / (0.021 + x)
Since NaOH is a strong base, it completely dissociates in water to produce hydroxide ions (OH-). Therefore, the concentration of OH- at the equivalence point is equal to the concentration of NaOH.
At the equivalence point, the concentration of NaOH will be:
0.40 M × (volume of NaOH added in L)
To find the volume of NaOH needed to reach the equivalence point, you need to calculate the number of moles of benzoic acid first.
Moles of benzoic acid = concentration × volume
0.42 M × 50.0 mL = 0.021 mol
Since 1 mol of benzoic acid reacts with 1 mol of NaOH, you need 0.021 moles of NaOH to reach the equivalence point.
Volume of NaOH needed to reach the equivalence point = 0.021 mol / 0.40 M = 0.0525 L = 52.5 mL
Now, you can calculate the concentration of OH- at the equivalence point:
Concentration of OH- = 0.40 M × 52.5 mL / 1000 mL = 0.021 M
Substitute the value of the concentration of OH- into the equation:
6.5 × 10^-5 = 0.021 - x / x
6.5 × 10^-5 = 0.021 - 0.021 / x
Solve for x, which gives the concentration of OH- at the equivalence point.
Once you have the concentration of OH-, you can calculate the pOH at the equivalence point using the formula:
pOH = -log[OH-]
Finally, you can calculate the pH at the equivalence point using the equation:
pH = 14 - pOH
Now, plug in the value of pOH into this equation to determine the pH at the equivalence point.