Question

If a weak acid is \(91 \%\) neutralized at \(\mathrm{pH} 5.7,\) what is the \(\mathrm{p} K^{\prime}\) of the acid?

Short answer

Answer: The pKa value of the weak acid is approximately 5.47.

Step-by-step solution

Determine the equilibrium constant expression for the weak acid

For any weak acid, HA, the dissociation equilibrium is given by: HA <=> H+ + A- The equilibrium constant expression, Ka, for the above reaction is: Ka = ([H+][A-])/([HA]) In terms of pKa, Ka = 10^(-pKa)

Express pH and pKa in terms of concentrations

We know that the pH is given as 5.7. We can use the definition of pH to relate it to the hydrogen ion concentration: pH = -log([H+]) Now, we can find the concentration of hydrogen ions [H+]: [H+] = 10^(-pH) = 10^(-5.7)

Find the ratio of [A-]/[HA]

Since the weak acid is 91% neutralized, we know that 91% of the acid has dissociated into its conjugate base. This means the ratio of [A-] to [HA] is: [A-]/[HA] = 0.91

Substitute the given values into the Ka expression

We already found the concentrations of [H+] and the ratio of [A-]/[HA]. We can now substitute these values into the Ka expression and solve for pKa: Ka = ([H+][A-])/([HA]) = 10^(-pKa) 10^(-pKa) = (10^(-5.7) * 0.91)

Solve for pKa

Now, we can solve for the pKa value by taking the negative logarithm of both sides of the equation: -pKa = log(10^(-5.7) * 0.91) pKa = -log(10^(-5.7) * 0.91) The pKa or pK' of the acid is approximately 5.47.