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Chapter 15: Problem 73

You have 75.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M}\) HA. After adding 30.0 \(\mathrm{mL}\) of 0.10 \(\mathrm{M} \mathrm{NaOH}\) , the \(\mathrm{pH}\) is \(5.50 .\) What is the \(K_{\mathrm{a}}\) value of \(\mathrm{HA}\) ?

Short Answer

Expert verified
The $K_{a}$ value of HA is approximately \(2.06 \times 10^{-5}\).

Step by step solution

01

Calculate the moles of HA and NaOH before the reaction

first, we will calculate the moles of HA and NaOH initially present using their concentrations and volumes: Moles of HA = Concentration of HA × Volume of HA = \(0.10\,\mathrm{M} \times 75.0\,\mathrm{mL}\) = \(0.0075\,\mathrm{mol}\) Moles of NaOH = Concentration of NaOH × Volume of NaOH = \(0.10\,\mathrm{M} \times 30.0\,\mathrm{mL}\) = \(0.0030\,\mathrm{mol}\)
02

Determine the limiting reagent, moles and concentrations of species after the reaction

Since NaOH is a strong base, it will completely react with the weak acid HA. So, the limiting reagent will be NaOH, and after the reaction: Moles of HA = \(0.0075\,\mathrm{mol} - 0.0030\,\mathrm{mol} = 0.0045\,\mathrm{mol}\) The reaction consumes all the moles of NaOH, forming an equal amount of the conjugate base A⁻. Moles of A⁻ = Moles of NaOH reacted = \(0.0030\,\mathrm{mol}\)
03

Calculate the concentrations of species after the reaction

Now, we calculate the new concentration of HA and A⁻ in the solution: Total volume of the solution = Initial volume of HA + Initial volume of NaOH = \(75.0\,\mathrm{mL} + 30.0\,\mathrm{mL}\) = \(105.0\,\mathrm{mL}\) Concentration of HA = Moles of HA / Total volume of the solution = \(0.0045\,\mathrm{mol} / 105.0\,\mathrm{mL}\) = \(0.0429\,\mathrm{M}\) Concentration of A⁻ = Moles of A⁻ / Total volume of the solution = \(0.0030\,\mathrm{mol} / 105.0\,\mathrm{mL}\) = \(0.0286\,\mathrm{M}\)
04

Use the pH value to find the concentration of hydrogen ions [H+]

The pH of the solution is 5.50, so we can find the concentration of H⁺ ions using: pH = -log\(_{10}\)([H⁺]) 5.50 = -log\(_{10}\)([H⁺]) [H⁺] = 10\(^{-5.50}\) = \(3.16 \times 10^{-6}\,\mathrm{M}\)
05

Use the equilibrium expression to calculate Ka

Using the equilibrium expression for the dissociation of HA: Ka = \(\frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]}\) Ka = \(\frac{(3.16 \times 10^{-6})(0.0286)}{0.0429}\) Ka = \(2.06 \times 10^{-5}\) So, the Ka value of HA is about \(2.06 \times 10^{-5}\).

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Most popular questions from this chapter

Another way to treat data from a pH titration is to graph the absolute value of the change in \(\mathrm{pH}\) per change in milliliters added versus milliliters added (\DeltapH/ \(\Delta \mathrm{mL}\) versus \(\mathrm{mL}\) added). Make this graph using your results from Exercise \(67 .\) What advantage might this method have over the traditional method for treating titration data?

Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HNO}_{3}\) with 0.100 $\mathrm{M} \mathrm{NaOH} .$

A student titrates an unknown weak acid, HA, to a pale pink phenolphthalein end point with 25.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{NaOH}\) . The student then adds 13.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{HCl}\) . The pH of the resulting solution is \(4.70 .\) How is the value of \(\mathrm{p} K_{2}\) for the unknown acid related to 4.70\(?\)

Malonic acid $\left(\mathrm{HO}_{2} \mathrm{CCH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)$ is a diprotic acid. In the titration of malonic acid with NaOH, stoichiometric points occur at \(\mathrm{pH}=3.9\) and $8.8 . \mathrm{A} 25.00-\mathrm{mL}$ sample of malonic acid of unknown concentration is titrated with 0.0984 \(\mathrm{M}\) NaOH, requiring 31.50 \(\mathrm{mL}\) of the NaOH solution to reach the phenolphthalein end point. Calculate the concentration of the initial malonic acid solution.

Consider the titration of 150.0 \(\mathrm{mL}\) of 0.100 $\mathrm{M} \mathrm{HI}\( by 0.250 \)\mathrm{M}\( \)\mathrm{NaOH}$ . a. Calculate the pH after 20.0 \(\mathrm{mL}\) of NaOH has been added. b. What volume of NaOH must be added so that the \(\mathrm{pH}=7.00 ?\)
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