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What is the \(\mathrm{pH}\) of a solution prepared by mixing \(60.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{H}_{3} \mathrm{PO}_{4}\) and \(1.00 \mathrm{~L}\) of \(0.100 \mathrm{M} \mathrm{LiOH} ?\)

Short Answer

Expert verified
The pH of the solution is approximately 1.60.

Step by step solution

01

Determine Moles of Each Reactant

First, calculate the moles of phosphoric acid (H鈧働O鈧) and lithium hydroxide (LiOH) using the concentration and volume. For H鈧働O鈧: Moles of H鈧働O鈧 = concentration 脳 volume = 1.00 M 脳 0.060 L = 0.060 mol. For LiOH: Moles of LiOH = concentration 脳 volume = 0.100 M 脳 1.00 L = 0.100 mol.
02

Write the Balanced Chemical Equation

The balanced chemical equation for the reaction between H鈧働O鈧 and LiOH is:\[ \mathrm{H}_{3}\mathrm{PO}_{4} + 3\mathrm{LiOH} \rightarrow \mathrm{Li}_3\mathrm{PO}_4 + 3\mathrm{H}_2\mathrm{O} \]One mole of H鈧働O鈧 reacts with three moles of LiOH.
03

Identify the Limiting Reagent

Calculate the moles of LiOH required to completely neutralize the H鈧働O鈧. 0.060 mol H鈧働O鈧 requires 3 脳 0.060 mol = 0.180 mol LiOH. Since we have only 0.100 mol of LiOH, LiOH is the limiting reagent.
04

Calculate Remaining H鈧働O鈧 and LiOH

Since LiOH is the limiting reagent, it will be completely consumed. For H鈧働O鈧: Remaining moles of H鈧働O鈧 = initial moles - moles reacted = 0.060 mol - (0.100 mol / 3) = 0.0267 mol. All LiOH is consumed.
05

Determine Resulting Solution Volume

The total volume of the resulting solution is the sum of the initial volumes: Total volume = 60.0 mL + 1000.0 mL = 1060.0 mL = 1.060 L.
06

Calculate Concentration of Remaining H鈧働O鈧

The concentration of H鈧働O鈧 remaining is given by:\[\text{Concentration of H}_{3}\text{PO}_{4} = \frac{\text{Moles of H}_{3}\text{PO}_{4}}{\text{Total Volume}} = \frac{0.0267 \text{ mol}}{1.060 \text{ L}} = 0.0252 \text{ M} \]
07

Calculate the pH of the Solution

Since H鈧働O鈧 is a weak acid that can donate up to three protons, and we've identified that there are excess H鈧働O鈧 molecules, we focus on the first ionization which has a higher influence on pH = -log鈧佲個[H鈦篯 Assuming full first ionization, [H鈦篯 鈮 [H鈧働O鈧 initial] = 0.0252 M. pH = -log(0.0252) 鈮 1.60.

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Key Concepts

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

Acid-Base Reaction
An acid-base reaction involves the exchange of protons between an acid and a base. In this exercise, we are looking at the reaction between phosphoric acid (H鈧働O鈧) and lithium hydroxide (LiOH). When an acid and a base react, they typically form water and a salt. These types of reactions are essential in various fields, including chemistry, biology, and environmental science.

Phosphoric acid is a triprotic acid, meaning it has three protons it can donate. On the other hand, lithium hydroxide is a strong base that can accept these protons. In our solution, when H鈧働O鈧 and LiOH are mixed, they undergo an acid-base reaction that primarily yields lithium phosphate (Li鈧働O鈧) and water. This process typically reduces the concentration of hydrogen ions in the solution, affecting the pH.
Limiting Reagent
In a chemical reaction, the limiting reagent is the reactant that is completely consumed first, stopping the reaction and determining the amount of product formed. In our example, to find the limiting reagent, you should compare the ratio of the moles required by the balanced equation with the moles available.

In the balanced equation for our exercise:
  • One mole of H鈧働O鈧 requires three moles of LiOH for complete reaction.
  • The initial moles of H鈧働O鈧 and LiOH are 0.060 mol and 0.100 mol respectively.
To neutralize all 0.060 mol of H鈧働O鈧, 0.180 mol of LiOH is needed. However, since only 0.100 mol of LiOH is present, LiOH is the limiting reagent. Recognizing the limiting reagent helps us know how much product can form and what will be in excess in the solution.
Neutralization Reaction
A neutralization reaction is a type of acid-base reaction where an acid and a base interact to form water and a salt. It's called neutralization because the acid and base neutralize each other's effects on the pH of the solution. In general, a strong acid and a strong base will neutralize each other to produce a pH close to 7. However, in this exercise, phosphoric acid (H鈧働O鈧) is a weak acid, meaning the situation is a bit different.

The balanced reaction between H鈧働O鈧 and LiOH is:
  • H鈧働O鈧 + 3LiOH 鈫 Li鈧働O鈧 + 3H鈧侽
This indicates the formation of lithium phosphate and water. Since LiOH is the limiting reagent, it gets entirely used up, and some phosphoric acid remains unreacted. The unreacted portion keeps the solution acidic, which is why calculating the pH of such a mixture involves considering the ionization of any excess acid.
Ionic Concentration
Ionic concentration refers to the concentration of ions in a solution. When calculating the pH of a solution resulting from an acid-base reaction, understanding the concentration of hydrogen ions ( [H鈦篯 ) is crucial as it directly impacts the pH.

In our case after the neutralization reaction has occurred, there is left-over H鈧働O鈧. This means that some hydrogen ions can dissociate, contributing to the solution's acidity. The concentration of this remaining acid is calculated based on the remaining moles and the total volume of the solution:
  • Concentration = 0.0267 mol of H鈧働O鈧 / 1.060 L = 0.0252 M
The first ionization of phosphoric acid will primarily affect the pH, given that it has the highest ionization constant. This ionic concentration is used in the equation pH = -log鈧佲個[H鈦篯 n to calculate the final pH of the solution.

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Most popular questions from this chapter

Calculate the concentrations of \(\mathrm{NH}_{4}^{+}\) and \(\mathrm{NH}_{3}\) and the \(\mathrm{pH}\) in a solution prepared by mixing \(20.0 \mathrm{~g}\) of \(\mathrm{NaOH}\) and \(0.500 \mathrm{~L}\) of \(1.5 \mathrm{M}\) \(\mathrm{NH}_{4} \mathrm{Cl}\). Assume that the volume remains constant.

Which of the following compounds are more soluble in acidic solution than in pure water? (a) \(\mathrm{AgCN}\) (b) \(\mathrm{PbI}_{2}\) (c) \(\mathrm{Al}(\mathrm{OH})_{3}\) (d) \(\mathrm{ZnS}\)

In an excess of \(\mathrm{NH}_{3}(a q), \mathrm{Cu}^{2+}\) ion forms a deep blue complex ion, \(\mathrm{Cu}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\), which has a formation constant \(K_{\mathrm{f}}=5.6 \times 10^{11} .\) Calculate the concentration of \(\mathrm{Cu}^{2+}\) in a solution prepared by adding \(5.0 \times 10^{-3} \mathrm{~mol}\) of \(\mathrm{CuSO}_{4}\) to \(0.500 \mathrm{~L}\) of \(0.40 \mathrm{M} \mathrm{NH}_{3}\).

Zinc hydroxide, \(\mathrm{Zn}(\mathrm{OH})_{2}\left(K_{\mathrm{sp}}=4.1 \times 10^{-17}\right)\), is nearly insoluble in water but is more soluble in strong base because \(\mathrm{Zn}^{2+}\) forms the soluble complex ion \(\mathrm{Zn}(\mathrm{OH})_{4}^{2-}\left(K_{\mathrm{f}}=3 \times 10^{15}\right)\) (a) What is the molar solubility of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in pure water? (You may ignore \(\mathrm{OH}^{-}\) from the self-dissociation of water.) (b) What is the \(\mathrm{pH}\) of the solution in part (a)? (c) What is the molar solubility of \(\mathrm{Zn}(\mathrm{OH})_{2}\) in \(0.10 \mathrm{M} \mathrm{NaOH}\) ?

The equivalence point was reached in titrations of three unknown bases at \(\mathrm{pH} 5.53\) (base A), \(4.11\) (base B), and \(6.00\) (base C). (a) Which is the strongest base? (b) Which is the weakest base?

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