Calculate the pH at the equivalence point if 25.00 mL of 0.010M barbituric acid (HC4H3N2O3) is titrated with 0.020M NaOH. (Ka barbituric acid = 1.0 x 10^-5)
If we call barituric acid, HB, the equation is
HB + NaOH ==> NaB + H2O, therefore, you can see that at the equivalence point we simply have the salt of barbituric acid so the pH will be determined by the hydrolysis of the salt.
You need to know the concn of the salt. Since the NaOH is twice the concn of the acid you know the volume of the base will be 12.5 mL so the concn of the salt at the equivalence point will be M = mols/L = 0.01 x (25/37.5) = 0.00667
........B^- + HOH ==> HB + OH^-
I...0.00667............0....0
C.......-x.............x....x
E...0.00667-x..........x....x
Kb for B^- = (Kw/Ka for HB) = (x)(x)/(0.00667-x)
Solve for x = (OH^-) and convert to pH.
Be careful with the algebra, you may need to use the quadratic equation since the concn of the salt is so small; i.e., 0.00667-x may not be equal to 0.00667
To calculate the pH at the equivalence point of this titration, we need to determine the amount of NaOH required to completely neutralize the barbituric acid and then calculate the resulting concentration of the conjugate base.
Step 1: Calculate the moles of barbituric acid (HC4H3N2O3) present in 25.00 mL of the 0.010 M solution.
Moles of HC4H3N2O3 = (0.010 M) x (0.025 L) = 0.00025 moles
Step 2: Determine the stoichiometry of the reaction between barbituric acid and NaOH.
The balanced equation for this reaction is:
HC4H3N2O3 + NaOH → NaC4H3N2O3 + H2O
From the balanced equation, we can see that one mole of barbituric acid reacts with one mole of NaOH, so the moles of NaOH required will be the same as the moles of barbituric acid.
Step 3: Calculate the volume of NaOH required to neutralize the barbituric acid.
Volume of NaOH = (0.00025 moles) / (0.020 M) = 0.0125 L = 12.5 mL
Step 4: Calculate the concentration of the conjugate base formed.
Moles of NaOH = (0.020 M) x (0.0125 L) = 0.00025 moles
Since the reaction is a 1:1 reaction, the concentration of the conjugate base (NaC4H3N2O3) formed will be:
Concentration of NaC4H3N2O3 = (0.00025 moles) / (0.0125 L) = 0.020 M
Step 5: Calculate the pOH at the equivalence point.
pOH = -log10(0.020) ≈ 1.70
Step 6: Calculate the pH at the equivalence point.
pH + pOH = 14 (for neutral solutions)
pH = 14 - 1.70 = 12.30
Therefore, the pH at the equivalence point is approximately 12.30.