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What is the \(\mathrm{pH}\) of a solution made by mixing \(0.30 \mathrm{~mol}\) \(\mathrm{NaOH}, 0.25 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{HPO}_{4},\) and \(0.20 \mathrm{~mol} \mathrm{H}_{3} \mathrm{PO}_{4}\) with water and diluting to \(1.00 \mathrm{~L} ?\)

Short Answer

Expert verified
The pH of the solution made by mixing 0.30 mol NaOH, 0.25 mol Na鈧侶PO鈧, and 0.20 mol H鈧働O鈧 with water and diluting to 1.00 L is approximately 1.06.

Step by step solution

01

Identify the species present in the solution

In this solution, there are three main species apart from water: 1. NaOH (strong base) 2. Na鈧侶PO鈧 (weak acid) 3. H鈧働O鈧 (stronger acid) NaOH and H鈧働O鈧 will react with each other completely because they are both strong electrolytes. The remaining weak acid (Na鈧侶PO鈧) will generate the pH of the solution.
02

Calculate the moles of unreacted species

First, we will find out how many moles of each species will remain after the reaction between NaOH and H鈧働O鈧. The limiting reagent in this reaction is H鈧働O鈧, as it has fewer moles than NaOH. Moles of H鈧働O鈧 reacted = \(0.20\, \cancel{mol} \cdot \frac{3 \, \cancel{mol}\, NaOH}{1\, \cancel{mol}\, H_3PO_4} = 0.60\, mol\, NaOH\) Moles of remaining NaOH = \(0.30\, mol\, NaOH - 0.60\, mol\, NaOH = -0.30\, mol\, NaOH\) Since we get a negative value for remaining NaOH, it means that NaOH is actually the limiting reagent in the reaction between NaOH and H鈧働O鈧. So, now we will recalculate the moles of the unreacted species based on NaOH being the limiting reagent. Moles of NaOH reacted = 0.30 mol Moles of H鈧働O鈧 reacted = \(0.30\, \cancel{mol}\, NaOH \cdot \frac{1 \, \cancel{mol}\, H_3PO_4}{3 \, \cancel{mol}\, NaOH} = 0.10\, mol\, H_3PO_4\) Moles of remaining H鈧働O鈧 = \(0.20\, mol\, H_3PO_4 - 0.10\, mol\, H_3PO_4 = 0.10\, mol\, H_3PO_4\) Now, we have 0.25 mol Na鈧侶PO鈧 and 0.10 mol H鈧働O鈧 in the solution.
03

Set up the chemical equilibrium equation

Since we have both Na鈧侶PO鈧 and H鈧働O鈧 in the solution, they will both contribute to the overall acidity of the solution. The equation that governs this equilibrium is as follows: H鈧働O鈧 + H鈧侽 鈬 H鈧侾O鈧勨伝 + H鈧僌鈦 K鈧 = \(\frac{[H_2PO_4^-][H_3O^+]}{[H_3PO_4]}\) (equation 1) Na鈧侶PO鈧 + H鈧侽 鈬 NaH鈧侾O鈧 + OH鈦 K_b = \(\frac{[NaH_2PO_4][OH^-]}{[Na_2HPO_4]}\) (equation 2) Kw = [H鈧僌鈦篯[OH鈦籡, where Kw is the ion-product constant of water = \(1.0 \times 10^{-14}\). Using these equations, we can find the concentration of H鈦 ions.
04

Find the concentration of H鈦 ions using the equilibrium constants

First, let's use equation 1 to find the concentration of H鈧僌鈦 ions contributed by H鈧働O鈧. Assume that x moles of H鈧働O鈧 dissociated to form H鈧侾O鈧勨伝 ions: K鈧 = \(\frac{x^2}{0.10-x}\) Since we are dealing with weak acid dissociation, it's safe to assume that x is very small compared to 0.10. So, we can approximate the equation as: K鈧 = \(\frac{x^2}{0.10}\) Now, let's use equation 2 to find the concentration of OH鈦 ions contributed by Na鈧侶PO鈧. Assume that y moles of Na鈧侶PO鈧 reacted to form NaH鈧侾O鈧 ions: K_b = \(\frac{y^2}{0.25-y}\) Again, assuming that y is very small compared to 0.25, we can approximate the equation as: K_b = \(\frac{y^2}{0.25}\) Since Kw = [H鈧僌鈦篯[OH鈦籡, we have the following equation: \(x \times y =1.0 \times 10^{-14}\)
05

Calculate the pH of the solution

Now that we have set up the equations, we can look up the constants K鈧, K_b, and Kw from a reference table. In our case, K鈧 = 7.5 脳 10鈦宦 and K_b = 1.6 脳 10鈦烩伔. Solving both the equilibrium equations, we get: \[x = \sqrt{0.10 \times 7.5 \times 10^{-3}} \approx 8.66 \times 10^{-2}\,M\text{ (concentration of H鈧僌鈦 ions)}\] \[y = \sqrt{0.25 \times 1.6 \times 10^{-7}} \approx 2.52 \times 10^{-4}\,M\text{ (concentration of OH鈦 ions)}\] Using the concentration of H鈧僌鈦 ions, we can now find the pH of the solution: pH = -log([H鈧僌鈦篯) = -log(8.66 脳 10鈦宦) 鈮 1.06 The pH of the solution made by mixing 0.30 mol NaOH, 0.25 mol Na鈧侶PO鈧, and 0.20 mol H鈧働O鈧 with water and diluting to 1.00 L is approximately 1.06.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Acid-Base Equilibrium
Understanding acid-base equilibrium is crucial when solving problems related to pH calculation in chemistry. At its core, this concept revolves around the idea that acids and bases in solution can react with each other to reach a state of balance, where the rate of the forward reaction equals the rate of the reverse reaction. In the mixture mentioned in the exercise, we are dealing with the weak acid Na鈧侶PO鈧 and the stronger acid H鈧働O鈧, both of which contribute to the solution's overall acidity.

In equilibrium terms, the presence of weak acid like Na鈧侶PO鈧 (which partially dissociates in water) and a strong acid like H鈧働O鈧 (which dissociates completely) creates a dynamic that directly impacts the concentration of hydronium ions (H鈧僌鈦) and thereby the pH of the solution. Moreover, the reaction between the strong base NaOH and the strong acid H鈧働O鈧 affects the concentration of remaining acid-base pairs and ultimately the equilibrium state. Utilizing the equilibrium constants (K鈧恈K_bc and KW) of the participating chemical species can help you calculate the final pH as illustrated in the step-by-step solution.
Chemical Species
The term 'chemical species' refers to the different forms of matter present in a chemical process, including elements, compounds, ions, or molecules. In the given pH calculation exercise, key species include a strong base (NaOH), a weak acid (Na鈧侶PO鈧), and a stronger acid (H鈧働O鈧).

It is important to identify these species correctly, as their individual properties, such as strength (as an acid or a base), impact how they interact and ultimately determine the pH of the solution. The strong base NaOH will completely dissociate in water, while the dissociation of Na鈧侶PO鈧 will be partial. Understanding the behavior of different chemical species in the mix allows for accurate stoichiometric calculations and proper setting-up of equilibrium expressions necessary for finding the pH.
Stoichiometry
Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It assists in calculations involving the moles of reactants and products. In this exercise, stoichiometry is used to determine the limiting reagent as well as the amounts of reactants and products after the reaction.

Initially, the exercise misidentified the limiting reagent, but a correct stoichiometric calculation revealed that NaOH was the limiting reagent. This adjustment is essential and impacts all subsequent calculations, including those related to the equilibrium of the solution. After establishing which reactants have reacted and which remain, stoichiometry then aids in setting up the proper molar ratios needed to find the pH of the solution. In essence, it underpins the entire process of determining the final acid-base equilibrium state of the chemical species in the solution.

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Most popular questions from this chapter

Assume that \(30.0 \mathrm{~mL}\) of a \(0.10 \mathrm{M}\) solution of a weak base \(\mathrm{B}\) that accepts one proton is titrated with a \(0.10 \mathrm{M}\) solution of the monoprotic strong acid HX. (a) How many moles of \(\mathrm{HX}\) have been added at the equivalence point? (b) What is the predominant form of \(\mathrm{B}\) at the equivalence point? (c) What factor determines the \(\mathrm{pH}\) at the equivalence point? (d) Which indicator, phenolphthalein or methyl red, is likely to be the better choice for this titration?

Consider the titration of \(30.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{NH}_{3}\) with \(0.025 \mathrm{M}\) HCl. Calculate the pH after the following volumes of titrant have been added: (a) \(0 \mathrm{~mL},(\mathbf{b}) 20.0 \mathrm{~mL},(\mathbf{c}) 59.0 \mathrm{~mL},(\mathrm{~d}) 60.0 \mathrm{~mL},\) (e) \(61.0 \mathrm{~mL}\) (f) \(65.0 \mathrm{~mL}\).

\(\begin{array}{llll}& \text { (a) If the molar solubility of } & \mathrm{CaF}_{2} & \text { at } & 35^{\circ} \mathrm{C} & \text { is }\end{array}\) \(1.24 \times 10^{-3} \mathrm{~mol} / \mathrm{L},\) what is \(K_{s p}\) at this temperature? (b) It is found that \(1.1 \times 10^{-2}\) of \(\mathrm{SrF}_{2}\) dissolves per \(100 \mathrm{~mL}\) of aqueous solution at \(25^{\circ} \mathrm{C} .\) Calculate the solubility product for \(\mathrm{SrF}_{2}\). (c) The \(K_{s p}\) of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) at \(25^{\circ} \mathrm{C}\) is \(6.0 \times 10^{-10} .\) What is the molar solubility of \(\mathrm{Ba}\left(\mathrm{IO}_{3}\right)_{2}\) ?

A buffer is prepared by adding \(20.0 \mathrm{~g}\) of sodium acetate \(\left(\mathrm{CH}_{3} \mathrm{COONa}\right)\) to \(500 \mathrm{~mL}\) of a \(0.150 \mathrm{M}\) acetic acid \(\left(\mathrm{CH}_{3} \mathrm{COOH}\right)\) solution. (a) Determine the \(\mathrm{pH}\) of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer.

A hypothetical weak acid, HA, was combined with \(\mathrm{NaOH}\) in the following proportions: \(0.20 \mathrm{~mol}\) of \(\mathrm{HA}, 0.080 \mathrm{~mol}\) of \(\mathrm{NaOH}\). The mixture was diluted to a total volume of \(1.0 \mathrm{~L}\) and the pH measured. (a) If \(\mathrm{pH}=4.80\), what is the \(\mathrm{p} K_{a}\) of the acid? (b) How many additional moles of \(\mathrm{NaOH}\) should be added to the solution to increase the \(\mathrm{pH}\) to \(5.00 ?\)

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