1. 40.0 mL of KOH solution added
2. 41.0 mL of KOH solution added

You will need to use your knowledge on salt solutions to recognize if either the cation or anion can react with water to impact the pH of the solution.

Since you have found the concentration of hydroxide, you will now need to recall the relationships that will allow you to calculate the pH and choose your approach.

“HBA” represents barbituric acid.

Use the given concentration and volume of HBA to calculate the moles of HBA present initially:

[latex]\begin{gathered} 40.0\mathrm{ml}\times\left(\frac{1\mathrm{~L}}{1000\mathrm{~ml}}\right)\times\left(\frac{0.100\mathrm{~mol}\;\mathrm{BA}}{1\mathrm{~L}}\right)=0.00400\mathrm{~mol}\;\mathrm{BA} \end{gathered}[/latex]

Use the added volume of KOH (40.0 mL) and the molarity (0.100 M) to calculate the moles of KOH added to this point:

[latex]\begin{gathered} 40.0\mathrm{ml}\times\left(\frac{1 \mathrm{~L}}{1000\mathrm{~ml}}\right)\times\left(\frac{0.100\mathrm{~mol}\;\mathrm{KOH}}{1\mathrm{~L}}\right)=0.00400\mathrm{~mol}\;\mathrm{KOH} \end{gathered}[/latex]

0.00400 moles KOH added – 0.00400 moles HBA present initially = 0.00000 moles KOH

This indicates moles of acid we started with equals the moles of base added, so we are at the equivalence point.

We look at the salt formed by the reaction of the acid and base and determine if it affects the pH of the solution. Break the salt into its cation and anion and react each with water.

Salt formed: K⁺ (aq) + BA^– (aq), or KBA (aq)

[latex]\begin{aligned} \mathrm{KBA}\,\mathrm{(s)}&\rightleftharpoons\mathrm{K}^{+}(\mathrm{aq})+\mathrm{BA}^{-}(\mathrm{aq}) \\ \mathrm{K}^{+}(\mathrm{aq})+\mathrm{H}_2\mathrm{O}\,(\ell)&\rightleftharpoons\text{No reaction}\\ \mathrm{BA}^{-}(\mathrm{aq})+\mathrm{H}_2\mathrm{O}\,(\ell)&\rightleftharpoons \mathrm{HBA}\,\mathrm{(aq)}+\mathrm{OH}^{-}(\text{aq}) \end{aligned}[/latex]

Since this is a basic salt, we need to determine the K_b of BA^– (aq) from the K_a of HBA.

[latex]\mathrm{HBA}\,\mathrm{K}_\mathrm{a}=9.8\times10^{−5}[/latex]

[latex]\begin{aligned} \mathrm{pK}_\mathrm{a}&=-\log(\mathrm{K}_\mathrm{a})=-\log(9.8 \times 10^{-5})=4.01 \\ \mathrm{pK}_\mathrm{b}&=14.00-\mathrm{pK}_\mathrm{a}=14.00-4.01=9.99 \\ \mathrm{K}_\mathrm{b}&=10^{\mathrm{-pK}_\mathrm{b}}=10^{-9.99}=1.0 \times 10^{-10} \end{aligned}[/latex]

Calculate the total solution volume by adding the initial volume of HBA to the volume of KOH (aq) added.

[latex]40.0\mathrm{~mL}+40.0\mathrm{~mL}=80.0\mathrm{~mL}[/latex]

[latex]\begin{gathered} \,[\mathrm{BA}^{-}]=\frac{0.0040 \mathrm{~mol}}{(80\mathrm{~ml})\times\left(\frac{1 \mathrm{~L}}{1000\mathrm{~ml}}\right)}=\frac{0.0040 \mathrm{~mol}}{0.080 \mathrm{~L}}=0.050\mathrm{~M} \end{gathered}[/latex]

Now, we do a weak base equilibrium calculation to determine the concentration of hydroxide in the solution.

[latex]\mathrm{BA}^-\mathrm{(aq)}+\mathrm{H}_2\mathrm{O}\,(\ell)\rightleftharpoons\mathrm{HBA}\,\mathrm{(aq)}+\mathrm{OH}^-\mathrm{(aq)}[/latex]

[latex]\begin{gathered} \mathrm{K}_\mathrm{b}=1.0 \times 10^{-10}=\frac{[\mathrm{HBA}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{BA}^{-}\right]}\\ \frac{[\mathrm{HBA}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{BA}^{-}\right]}=\frac{(\mathrm{x})\left(1.0 \times 10^{-7}+\mathrm{x}\right)}{0.050-\mathrm{x}}=\frac{\mathrm{(x)(x)}}{0.050} \end{gathered}[/latex]

[latex]\begin{aligned} \frac{\mathrm{x}^2}{0.050}&=1.0 \times10^{-10} \\ \mathrm{x}^2&=1.0 \times 10^{-10}(0.050) \\ \mathrm{x}^2&=5.0\times 10^{-12} \\ \sqrt{\mathrm{x}^2}&=\sqrt{5.0 \times 10^{-12}} \\ [\mathrm{OH}^{-}]=\mathrm{x}&=2.2\times 10^{-6} \end{aligned}[/latex]

Since we want pH, we need to use the hydroxide concentration to calculate it in 2 steps.

[latex]\begin{aligned} \mathrm{pOH}&=-\log[\mathrm{OH}^-]=-\log(2.2\times10^{-6})=5.66\\ \mathrm{pH}&=14.000-\mathrm{pOH}= 14.000-5.66=8.34 \end{aligned}[/latex]

“HBA” represents barbituric acid.

Calculate the moles of HBA:

Moles of HBA present initially:

[latex]\begin{gathered} 40.0\mathrm{ml}\times\left(\frac{1\mathrm{~L}}{1000\mathrm{~ml}}\right)\times\left(\frac{0.100\mathrm{~mol}\;\mathrm{BA}}{1\mathrm{~L}}\right)=0.00400\mathrm{~mol}\;\mathrm{BA} \end{gathered}[/latex]

Calculate the moles of KOH:

Moles of KOH added

[latex]\begin{gathered} 41.0\mathrm{~ml}\times\left(\frac{1\mathrm{~L}}{1000 \mathrm{~ml}}\right) \times\left(\frac{0.100 \mathrm{~mol}\;\mathrm{KOH}}{1 \mathrm{~L}}\right)=0.00410 \mathrm{~mol}\,\mathrm{KOH} \end{gathered}[/latex]

Find the moles of KOH left after the reaction:

0.00410 moles KOH added – 0.00400 moles BA present initially = 0.00010 moles KOH

This means there is excess KOH, and we have gone past the equivalence point. We need to calculate the concentration of the excess hydroxide.

Calculate the total solution volume by adding inital volume of HBA to the volume of KOH solution added:

[latex]41.0\mathrm{~mL}+40.0\mathrm{~mL}=81.0\mathrm{~mL}[/latex]

Use the moles of excess hydroxide and the total volume to calculate hydroxide concentration:

[latex]\begin{gathered} \left[\mathrm{OH}^{-}\right]=\frac{0.00010 \mathrm{~mol}\;\mathrm{KOH}}{(81.0 \mathrm{~ml}) \times\left(\frac{1 \mathrm{~L}}{1000 \mathrm{~ml}}\right)}=1.2 \times 10^{-3} \mathrm{~M} \end{gathered}[/latex]

To calculate pH:

[latex]\begin{aligned} \mathrm{pOH}&=-\log[\mathrm{OH}^-]=-\log(2.2\times10^{-3})=2.92\\ \mathrm{pH}&=14.000-\mathrm{pOH}=14.000-2.92=11.08 \end{aligned}[/latex]