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Chapter 15: Problem 133

Calculate the pH of a solution prepared by mixing \(250 . \mathrm{mL}\) of 0.174 \(\mathrm{m}\) aqueous \(\mathrm{HF}\) (density \(=1.10 \mathrm{g} / \mathrm{mL} )\) with 38.7 \(\mathrm{g}\) of an aqueous solution that is 1.50\(\% \mathrm{NaOH}\) by mass (density \(=\) 1.02 $\mathrm{g} / \mathrm{mL} ) .\left(K_{\mathrm{a}} \text { for } \mathrm{HF}=7.2 \times 10^{-4} .\right)$

Short Answer

Expert verified
The pH of the solution prepared by mixing 250 mL of 0.174 M HF with 38.7 g of 1.50% NaOH solution is approximately 2.57.

Step by step solution

01

Calculate the moles of HF in the solution

To determine the moles of HF in the initial solution, we can use the formula: moles of HF = volume of solution (in L) × molarity of solution moles of HF = \(0.250 L \times 0.174 \frac{mol}{L}\) moles of HF = 0.0435 mol
02

Calculate the moles of NaOH in the solution

To determine the moles of NaOH in the initial solution, we can first find the mass of NaOH present in the solution, and then convert it to moles using its molar mass. Mass of NaOH = mass of solution × percentage of NaOH by mass Mass of NaOH = \(38.7 g \times 0.015\) Mass of NaOH = 0.5805 g Moles of NaOH = Mass of NaOH / Molar mass of NaOH Moles of NaOH = \(0.5805 g \div 40.00 \frac{g}{mol}\) Moles of NaOH = 0.01451 mol
03

Perform the reaction between HF and NaOH

The reaction between HF and NaOH can be represented as follows: HF(aq) + NaOH(aq) → NaF(aq) + H2O(l) Since the moles of NaOH is less than that of HF, NaOH will act as the limiting reagent and will completely react with HF. The moles of HF and NaOH left after the reaction are: Moles of HF left = moles of HF initially - moles of NaOH reacted Moles of HF left = 0.0435 mol - 0.01451 mol Moles of HF left = 0.02899 mol Moles of NaOH left = 0 mol (since it completely reacted)
04

Calculate the Ka expression for HF in the final solution

The dissociation of HF can be represented as: HF(aq) + H2O(l) ↔ H3O+(aq) + F-(aq) The Ka expression for HF is: Ka = \(\frac{[H_{3}O^{+}][F^{-}]}{[HF]}\) Since NaOH is completely reacted, only the remaining HF will dissociate to produce H3O+ and F- ions. Let x be the concentration of H3O+ ions produced. Then the concentration of F- ions will also be x. As the total volume of the solution is 0.250 L + 0.038 L = 0.288 L, we have: Initial concentration of HF = \(\frac{0.02899 \, mol}{0.288 \, L}\) = 0.1007 M Ka = \(\frac{x \times x}{0.1007 - x} = 7.2 \times 10^{-4}\)
05

Solve for x and calculate pH

Solve the equation for x, which is the concentration of H3O+ ions: \(x^{2} = 7.2 \times 10^{-4} \times (0.1007 - x)\) Since the value of x is small compared to 0.1007, the equation can be simplified to: \(x^{2} = 7.2 \times 10^{-4} \times 0.1007\) Solve for x: x = 0.00271 The pH of the solution can be calculated using the formula: pH = -log[H3O+] pH = -log(0.00271) pH ≈ 2.57 The pH of the solution prepared by mixing 250 mL of 0.174 M HF with 38.7 g of 1.50% NaOH solution is approximately 2.57.

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Most popular questions from this chapter

Calculate the number of moles of \(\mathrm{HCl}(g)\) that must be added to 1.0 \(\mathrm{L}\) of 1.0 $\mathrm{M} \mathrm{NaC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}$ to produce a solution buffered at each pH. $$ \text{(a)}\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} \quad \text { b. } \mathrm{pH}=4.20 \quad \text { c. } \mathrm{pH}=5.00 $$

Consider the titration of 100.0 \(\mathrm{mL}\) of 0.10 $\mathrm{M} \mathrm{H}_{3} \mathrm{AsO}_{4}\( by 0.10 \)M \mathrm{NaOH}$ . What are the major species present at 50.0 \(\mathrm{mL}\) of \(\mathrm{NaOH}\) added? How would you calculate the pH at this point? Answer the same questions for 150.0 \(\mathrm{mL}\) of NaOH added. At what volume of NaOH added does pH $=\mathrm{p} K_{\mathrm{a}_{\mathrm{i}}} ?$

You have a solution of the weak acid HA and add some of the salt NaA to it. What are the major species in the solution? What do you need to know to calculate the pH of the solution, and how would you use this information? How does the pH of the solution of just the HA compare with that of the final mixture? Explain

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 \(=7.40\) $$ \mathrm{H}_{2} \mathrm{CO}_{3}(a q) \rightleftharpoons \mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=4.3 \times 10^{-7} $$

Acid-base indicators mark the end point of titrations by "magically" turning a different color. Explain the "magic" behind acid-base indicators.
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