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Chapter 15: Problem 42

A buffered solution is made by adding 50.0 $\mathrm{g} \mathrm{NH}_{4} \mathrm{Cl}\( to 1.00 \)\mathrm{L}\( of a \)0.75-\mathrm{M}$ solution of \(\mathrm{NH}_{3}\) . Calculate the \(\mathrm{pH}\) of the final solution. (Assume no volume change.)

Short Answer

Expert verified
The pH of the final buffered solution is 9.14 after adding 50.0 g of NH4Cl to 1.00 L of a 0.75 M solution of NH3, calculated using the Henderson-Hasselbalch equation.

Step by step solution

01

Identify the relevant equation

The acid dissociation constant (Ka) for NH4+ reaction with water is given by: \(K_\mathrm{a}=\frac{\ce{[H+][NH3]}}{\ce{[NH4+]}}\) The Henderson-Hasselbalch equation is used to relate the pH, pKa, and concentrations of a weak acid ([NH4+]) and its conjugate base ([NH3]) in a buffer solution: \(pH=pK_\mathrm{a} + \log{\frac{[\ce{NH3}]}{[\ce{NH4+}]}}\)
02

Determine the concentration of NH4Cl solution

First, we need to find the concentration of NH4Cl in the solution. We have the mass (50.0g) and the volume of NH3 solution (1L). To calculate the concentration, we need to convert the mass of NH4Cl into moles and divide by the volume. Molar mass of NH4Cl (g/mol): \( M_\mathrm{NH4Cl} = 14.01 \text{(N)} + 1.01 (\text{H}) * 4 + 35.45\text{(Cl)} \) \( M_\mathrm{NH4Cl} = 53.49\, \mathrm{g/mol} \) To find the moles of NH4Cl: Moles of NH4Cl = 50.0 g / 53.49 g/mol Moles of NH4Cl = 0.935 mol Finally, we can calculate the concentration of NH4Cl: Concentration of NH4Cl (M) = moles / volume Concentration of NH4Cl = 0.935 mol / 1.00 L Concentration of NH4Cl = 0.935 M
03

Find the pKa using the given Ka

Next, we need to find the pKa value. The pKa value is the negative logarithm of the Ka value. Given the Ka value for NH4+ is 5.56E-10, we can find the pKa as: \(pK_\mathrm{a} = -\log(5.56 \times 10^{-10})\) \(pK_\mathrm{a} = 9.25\)
04

Calculate the pH using the Henderson-Hasselbalch equation

Now, we can use the Henderson-Hasselbalch equation to find the pH of the buffered solution: \(pH = pK_\mathrm{a} + \log{\frac{[\ce{NH3}]}{[\ce{NH4+}]}}\) Using the concentrations of NH3 (0.75 M) and NH4+ (0.935 M) and the pKa value (9.25), we get: \(pH = 9.25 + \log{\frac{0.75}{0.935}}\) \(pH = 9.25 - 0.112\) \(pH = 9.14\) The pH of the final solution is 9.14.

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Most popular questions from this chapter

Mixing together solutions of acetic acid and sodium hydroxide can make a buffered solution. Explain. How does the amount of each solution added change the effectiveness of the buffer?

Carbonate buffers are important in regulating the pH of blood at \(7.40 .\) If the carbonic acid concentration in a sample of blood is 0.0012 M, determine the bicarbonate ion concentration required to buffer the pH of blood at pH \(=7.40\) $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=4.3 \times 10^{-7} $$

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intra- cellular fluids at pH values generally between 7.1 and \(7.2 .\) a. What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) inintracellular fluid at \(\mathrm{pH}=7.15 ?\) $$ \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8} $$ b. Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the pH of intracellular fluid? $$ \mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{2}=7.5 \times 10^{-3} $$

Consider the titration of 80.0 \(\mathrm{mL}\) of 0.100 $\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\( by 0.400 \)\mathrm{M} \mathrm{HCl}$ . Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of HCl have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}} \\ {\text { c. } 30.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}}\end{array} $$

A certain acetic acid solution has \(\mathrm{pH}=2.68\) . Calculate the volume of 0.0975 \(\mathrm{M} \mathrm{KOH}\) required to reach the equivalence point in the titration of 25.0 \(\mathrm{mL}\) of the acetic acid solution.
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