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Chapter 17: Problem 41

Calculate the \(\mathrm{pH}\) at the points in the titration of \(25.00 \mathrm{mL}\) of \(0.132 \mathrm{M} \mathrm{HNO}_{2}\) when (a) \(10.00 \mathrm{mL}\) and (b) 20.00 mL of 0.116 M \(\mathrm{HNO}_{2}\) when \((\mathrm{a}) 10.00 \mathrm{mL}\) and (b) \(20.00 \mathrm{mL}\) of \(0.116 \mathrm{M}\) NaOH have been added. For \(\mathrm{HNO}_{2}, K_{\mathrm{a}}=7.2 \times 10^{-4}\) \(\mathrm{HNO}_{2}+\mathrm{OH}^{-} \longrightarrow \mathrm{H}_{2} \mathrm{O}+\mathrm{NO}_{2}^{-}\)

Short Answer

Expert verified
The pH after adding 10.00 mL NaOH is calculated to be approximately 2.43, and after adding 20.00 mL NaOH it is approximated to be about 2.78. The exact values depend on the Ka of HNO2 and may be slightly different due to rounding.

Step by step solution

01

Calculations for 10.00 mL NaOH

For (a) First, find the moles of HNO2 and NaOH. Moles of HNO2 = volume (L) x concentration (M) = 0.025 L x 0.132 M = 0.0033 mol. Moles of NaOH = volume (L) x concentration (M) = 0.01 L x 0.116 M = 0.00116 mol. Since the moles of HNO2 are more than the moles of NaOH, some HNO2 will remain after the reaction. Moles of remaining HNO2 = moles of initial HNO2 - moles of NaOH = 0.0033 mol - 0.00116 mol = 0.00214 mol.
02

pH calculation for 10.00 mL NaOH

Now use the Ka expression to find the concentration of H3O+. Ka = [H3O+][NO2-] / [HNO2]. H3O+ and NO2- are produced by the ionization of HNO2, and because we start with 0 H3O+ and NO2-, the concentrations of H3O+ and NO2- at equilibrium are equal. So, set Ka = x^2 / ([HNO2] - x), where x is the concentration of H3O+ and NO2-, and [HNO2] is the concentration of the remaining HNO2. Solve for x, and then calculate pH = -log(x).
03

Calculations for 20.00 mL NaOH

For (b) Repeat the process for 20.00 mL NaOH. Moles of NaOH = 0.02 L x 0.116 M = 0.00232 mol. There are still more moles of HNO2, so again, some will remain after the reaction. Moles of remaining HNO2 = 0.0033 mol - 0.00232 mol = 0.00098 mol.
04

pH calculation for 20.00 mL NaOH

Now calculate the pH as in step 2. The answer will be higher than before, because adding more NaOH increases the pH.

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Most popular questions from this chapter

Sodium ammonium hydrogen phosphate, \(\mathrm{NaNH}_{4}\) \(\mathrm{HPO}_{4},\) is a salt in which one of the ionizable \(\mathrm{H}\) atoms of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is replaced by \(\mathrm{Na}^{+},\) another is replaced by \(\mathrm{NH}_{4}^{+},\) and the third remains in the anion \(\mathrm{HPO}_{4}^{2-}\) Calculate the \(\mathrm{pH}\) of \(0.100 \mathrm{M} \mathrm{NaNH}_{4} \mathrm{HPO}_{4}(\mathrm{aq})\) [Hint: You can use the general method introduced on page \(720 .\) First, identify all the species that could be present and the equilibria involving these species. Then identify the two equilibrium expressions that will predominate and eliminate all the species whose concentrations are likely to be negligible. At that point, only a few algebraic manipulations are required.]

Rather than calculate the \(\mathrm{pH}\) for different volumes of titrant, a titration curve can be established by calculating the volume of titrant required to reach certain \(\mathrm{pH}\) values. Determine the volumes of \(0.100 \mathrm{M} \mathrm{NaOH}\) required to reach the following pH values in the titration of \(20.00 \mathrm{mL}\) of \(0.150 \mathrm{M} \mathrm{HCl}: \mathrm{pH}=\) (a) 2.00 (b) \(3.50 ;\) (c) \(5.00 ;\) (d) \(10.50 ;\) (e) \(12.00 .\) Then plot the titration curve.

Consider a solution containing two weak monoprotic acids with dissociation constants \(K_{\mathrm{HA}}\) and \(K_{\mathrm{HB}}\). Find the charge balance equation for this system, and use it to derive an expression that gives the concentration of \(\mathrm{H}_{3} \mathrm{O}^{+}\) as a function of the concentrations of \(\mathrm{HA}\) and HB and the various constants.

A 25.00 -mL sample of \(0.0100 \mathrm{M} \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{COOH}\left(\mathrm{K}_{\mathrm{a}}=\right.\) \(\left.6.3 \times 10^{-5}\right)\) is titrated with \(0.0100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) Calculate the \(\mathrm{pH}\) (a) of the initial acid solution; (b) after the addition of 6.25 mL of \(0.0100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) (c) at the equivalence point; (d) after the addition of a total of \(15.00 \mathrm{mL}\) of \(0.0100 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\)

Explain why the volume of \(0.100 \mathrm{M} \mathrm{NaOH}\) required to reach the equivalence point in the titration of \(25.00 \mathrm{mL}\) of \(0.100 \mathrm{M}\) HA is the same regardless of whether HA is a strong or a weak acid, yet the \(\mathrm{pH}\) at the equivalence point is not the same.
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