Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 14: Problem 17
What is the \(\mathrm{pH}\) of the solution that results when \(0.093 \mathrm{~g}\) of \(\mathrm{Mg}(\mathrm{OH})_{2}\) is mixed with (a) \(75.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{HCl}\) ? (b) \(100.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{HCl}\) ? (c) \(15.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{HCl}\) ? (d) \(30.0 \mathrm{~mL}\) of \(0.0500 \mathrm{M} \mathrm{MgCl}_{2}\) ?
Short Answer
Expert verified
(a) pH 鈮� 2.23; (b) pH 鈮� 1.82; (c) pH 鈮� 10.96; (d) pH 鈮� 10.5.
Step by step solution
01
Calculate moles of Mg(OH)2
To find the number of moles of \( \mathrm{Mg(OH)_2} \), use the formula \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). The molar mass of \( \mathrm{Mg(OH)_2} \) is approximately \( 58.32 \, \mathrm{g/mol} \). Thus, \( \text{moles of } \mathrm{Mg(OH)_2} = \frac{0.093}{58.32} \approx 0.0016 \, \mathrm{mol} \).
02
Calculate moles of HCl
For each scenario, calculate the moles of \( \mathrm{HCl} \) using the formula \( \text{moles} = \text{Molarity} \times \text{Volume} \). The molarity is \( 0.0500 \, M \).- (a) For 75.0 mL: \( 0.0500 \, M \times 0.0750 \, L = 0.00375 \, \mathrm{mol} \)- (b) For 100.0 mL: \( 0.0500 \, M \times 0.100 \, L = 0.0050 \, \mathrm{mol} \)- (c) For 15.0 mL: \( 0.0500 \, M \times 0.0150 \, L = 0.00075 \, \mathrm{mol} \)
03
Determine excess reactant and calculate resulting pH (Scenarios a-c)
\( \mathrm{Mg(OH)_2} \) reacts with \( \mathrm{HCl} \) as: \[ \mathrm{Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O} \]**Part (a):**- \( 0.0016 \, \mathrm{mol} \mathrm{Mg(OH)_2} \) requires \( 0.0032 \, \mathrm{mol} \) of \( \mathrm{HCl} \) and reacts fully with \( 0.00375 \, \mathrm{mol} \) of \( \mathrm{HCl} \).- \( 0.00375 - 0.0032 = 0.00055 \, \mathrm{mol} \) \( \mathrm{HCl} \) excess.- \([\mathrm{H^+}] = \frac{0.00055}{0.0932} = 0.0059 \, M\) (use total volume for molarity).- \( \mathrm{pH} = -\log[\mathrm{H^+}] \approx 2.23 \).**Part (b):**- \( 0.0016 \, \mathrm{mol} \mathrm{Mg(OH)_2} \) requires \( 0.0032 \, \mathrm{mol} \mathrm{HCl} \) and reacts with \( 0.0050 \, \mathrm{mol} \mathrm{HCl} \).- \( 0.0050 - 0.0032 = 0.0018 \, \mathrm{mol} \mathrm{HCl} \) excess.- \([\mathrm{H^+}] = \frac{0.0018}{0.1192} = 0.0151 \, M\).- \( \mathrm{pH} = -\log[\mathrm{H^+}] \approx 1.82 \).**Part (c):**- \( 0.0016 \, \mathrm{mol} \mathrm{Mg(OH)_2} \) doesn't fully react as \( \mathrm{HCl} \) available is 0.00075 mol.- \( 0.0016 - 0.00075 \cdot 2 = 0.0001 \, \mathrm{mol} \mathrm{OH}^- \) excess.- \([\mathrm{OH}^-] = \frac{0.0001}{0.1082} = 0.00092 \, M\).- \( \mathrm{pOH} = -\log[\mathrm{OH}^-] \approx 3.04 \) and \( \mathrm{pH} = 14 - \mathrm{pOH} = 10.96 \).
04
Calculate moles of MgCl2 and its effect on pH (Scenario d)
\( \mathrm{MgCl_2} \) doesn't interact with \( \mathrm{Mg(OH)_2} \) since \( \mathrm{MgCl_2} \) is a neutral salt and doesn't produce \( \mathrm{H^+} \) ions.- No dissolution affects \( \mathrm{pH} \) when mixed with \( \mathrm{Mg(OH)}_2 \), as it doesn鈥檛 react with it.-\( \mathrm{pH} \) remains basic due to \( \mathrm{Mg(OH)_2} \).- Given solid \( \mathrm{Mg(OH)_2} \) excess, assume a basic \( \mathrm{pH} \approx 10.5 \).
05
Final Step: Summarize the Answers
Based on calculations:- (a) The \( \mathrm{pH} \) is approximately 2.23.- (b) The \( \mathrm{pH} \) is approximately 1.82.- (c) The \( \mathrm{pH} \) is approximately 10.96.- (d) The solution remains basic due to excess \( \mathrm{Mg(OH)_2} \), \( \mathrm{pH} \approx 10.5 \).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
pH Calculations
Understanding how to calculate the pH of different solutions is crucial in acid-base chemistry. The pH scale measures how acidic or basic a solution is. It ranges from 0 to 14, where a pH less than 7 is acidic, a pH of 7 is neutral, and a pH greater than 7 is basic. To calculate the pH, typically you determine the concentration of hydrogen ions (H^+) in the solution. The formula for calculating pH is:\[\mathrm{pH} = -\log[\mathrm{H^+}]\]For example, in scenario (a), the excess HCl provides an H^+ concentration of 0.0059 M, resulting in a calculated pH of approximately 2.23. This tells us that the solution is quite acidic. On the other hand, scenarios involving Mg(OH)_2, like (c) and (d), can leave excess OH^{-} ions, which instead requires calculation of pOH using:\[\mathrm{pOH} = -\log[\mathrm{OH^-}]\]You then find the pH by subtracting the pOH from 14. Hence, knowing how to determine both H^+ and OH^{-} concentrations is essential.
Stoichiometry
Stoichiometry helps us understand the quantitative relationships between reactants and products in a chemical reaction. For reactions involving acids and bases, stoichiometry allows us to determine how much acid is needed to react with a given quantity of base, and vice versa.In our exercise, you can see the stoichiometry play out with the neutralization reaction:\[ \mathrm{Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O} \]Here, one mole of Mg(OH)_2 reacts with two moles of HCl. This stoichiometric relationship is essential when calculating the excess of either reactant, as it defines the amounts needed for a complete reaction. For instance, in part (b), the 0.0016 mol of Mg(OH)_2 requires 0.0032 mol of HCl. Since 0.0050 mol of HCl is present, an excess of HCl exists, affecting the resulting H^+ concentration and hence the pH. Understanding these quantitative relationships is vital in any chemical calculations involving reactions.
Chemical Reactions
A chemical reaction is a process where reactants are transformed into products. In acid-base chemistry, these often involve neutralization reactions, where an acid and a base react to form water and a salt.In our exercise, Mg(OH)_2 and HCl undergo such a reaction:\[ \mathrm{Mg(OH)_2 + 2HCl \rightarrow MgCl_2 + 2H_2O} \]The reaction involves the transfer of protons from the acid (HCl) to the hydroxide from the base (Mg(OH)_2), forming water. This proton transfer is crucial, as it determines the formation of the neutral products. These types of reactions are often complete, meaning they go to completion with no remaining reactants, unless one is in excess. For instance, MgCl2 in part (d) does not react with Mg(OH)_2 because it is a neutral salt, illustrating how specific reactions rely on the chemical nature of interacting substances.
Solution Chemistry
Solution chemistry involves understanding how substances dissolve and interact in a solvent, often water. Concentration plays a key role here, defined as the amount of solute per volume of solvent, commonly expressed in molarity (M, or mol/L).In our context, the mixing involves specific molar concentrations of HCl and MgCl2. For example, scenario (b) uses 100.0 mL of 0.0500 M HCl. From this, we calculate the moles of HCl using:\[ \text{Moles}=\text{Molarity} \times \text{Volume} \]Which gives us an HCl concentration that, when mixed with Mg(OH)_2, determines the extent of reaction and the remaining excess of ions.Additionally, dilution affects concentrations after mixing. As the volume of the solution increases, the concentration of remaining reactants or products changes. These concepts help us predict the outcome of reactions in solution, including effects on pH, as each component's concentration is a decisive factor.
