Problem 82 - Chemistry The Central Science In Si Units

Question

(a) How many milliliters of \(0.120 \mathrm{M} \mathrm{HCl}\) are needed to completely neutralize \(50.0 \mathrm{~mL}\) of \(0.101 \mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) solution? (b) How many milliliters of \(0.125 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}\) are needed to neutralize \(0.200 \mathrm{~g}\) of \(\mathrm{NaOH} ?\) (c) If \(55.8 \mathrm{~mL}\) of a \(\mathrm{BaCl}_{2}\) solution is needed to precipitate all the sulfate ion in a \(752-\mathrm{mg} \mathrm{sam}-\) ple of \(\mathrm{Na}_{2} \mathrm{SO}_{4},\) what is the molarity of the \(\mathrm{BaCl}_{2}\) solution? (d) If \(42.7 \mathrm{~mL}\) of \(0.208 \mathrm{M}\) HCl solution is needed to neutralize a solution of \(\mathrm{Ca}(\mathrm{OH})_{2},\) how many grams of \(\mathrm{Ca}(\mathrm{OH})_{2}\) must be in the solution?

Step-by-Step Solution

Answer

(a) 84.2 mL of 0.120 M HCl. (b) 20 mL of 0.125 M H2SO4. (c) 0.0654 M BaCl2. (d) 0.329 g of Ca(OH)2.

Step 1: Determine the Balanced Chemical Equation for Reaction (a)

For the reaction between HCl and \(Ba(OH)_2\), the balanced equation is: \[ 2 \mathrm{HCl} + \mathrm{Ba(OH)}_2 \rightarrow 2 \mathrm{H}_2\mathrm{O} + \mathrm{BaCl}_2 \] This indicates that 2 moles of HCl react with 1 mole of \(Ba(OH)_2\).

Step 2: Calculate the Moles of \(Ba(OH)_2\) (Part a)

First, find the moles of \(Ba(OH)_2\) using the concentration and volume: \[ \text{Moles of } Ba(OH)_2 = 0.101 \mathrm{~M} \times 0.0500 \mathrm{~L} = 0.00505 \mathrm{~moles} \]

Step 3: Calculate the Required Moles of HCl (Part a)

Using the stoichiometry from Step 1, calculate the moles of HCl required: \[ \text{Moles of HCl} = 2 \times 0.00505 = 0.0101 \mathrm{~moles} \]

Step 4: Calculate the Volume of \(0.120 \mathrm{M} \) HCl Needed (Part a)

Using the moles of HCl from Step 3: \[ V = \frac{0.0101 \mathrm{~moles}}{0.120 \mathrm{~M}} = 0.0842 \mathrm{~L} = 84.2 \mathrm{~mL} \]

Step 5: Determine the Reaction for Question (b)

The reaction between \(H_2SO_4\) and \(NaOH\) is: \[ \mathrm{H}_2\mathrm{SO}_4 + 2 \mathrm{NaOH} \rightarrow \mathrm{Na}_2\mathrm{SO}_4 + 2 \mathrm{H}_2\mathrm{O} \] Each mole of \(H_2SO_4\) reacts with 2 moles of \(NaOH\).

Step 6: Calculate the Moles of NaOH (Part b)

Calculate the moles of \(NaOH\): \[ \text{Molar mass of NaOH} = 40 \mathrm{~g/mol} \] \[ \text{Moles of NaOH} = \frac{0.200 \mathrm{~g}}{40 \mathrm{~g/mol}} = 0.005 \mathrm{~moles} \]

Step 7: Calculate the Moles of H2SO4 Needed (Part b)

Since 1 mole of \(H_2SO_4\) reacts with 2 moles of NaOH: \[ \text{Moles of } H_2SO_4 = \frac{0.005 \text{ moles of } NaOH}{2} = 0.0025 \mathrm{~moles} \]

Step 8: Calculate the Volume of \(0.125 \mathrm{M} \) \(H_2SO_4\) Needed (Part b)

Using the moles of \(H_2SO_4\) from Step 7: \[ V = \frac{0.0025 \mathrm{~moles}}{0.125 \mathrm{~M}} = 0.02 \mathrm{~L} = 20 \mathrm{~mL} \]

Step 9: Determine the Balanced Equation for Part (c)

For the reaction between \(BaCl_2\) and \(Na_2SO_4\), the balanced equation is: \[ \mathrm{BaCl}_2 + \mathrm{Na}_2\mathrm{SO}_4 \rightarrow 2 \mathrm{NaCl} + \mathrm{BaSO}_4 \] 1 mole of \(BaCl_2\) reacts with 1 mole of \(Na_2SO_4\).

Step 10: Calculate the Moles of \(Na_2SO_4\) in Sample (Part c)

Determine the moles of \(Na_2SO_4\) in the sample: \[ \text{Molar mass of } Na_2SO_4 = 142 + 64 = 142 + 64 \] \[ 142 + 64 = 206 \mathrm{~g/mol} \] \[ \text{Moles of } Na_2SO_4 = \frac{752 \mathrm{~mg}}{206 \mathrm{~g/mol}} = 0.00365 \mathrm{~moles} \]

Step 11: Calculate Molarity of \(BaCl_2\) Solution Needed (Part c)

The moles of \(BaCl_2\) are equal to moles of \(Na_2SO_4\), thus: \[ C = \frac{0.00365 \mathrm{~moles}}{0.0558 \mathrm{~L}} = 0.0654 \mathrm{~M} \]

Step 12: Determine the Reaction for Part (d)

For \(HCl\) and \(Ca(OH)_2\): \[ 2 \mathrm{HCl} + \mathrm{Ca(OH)}_2 \rightarrow 2 \mathrm{H}_2\mathrm{O} + \mathrm{CaCl}_2 \] 2 moles of HCl react with 1 mole of \(Ca(OH)_2\).

Step 13: Calculate Moles of HCl Used (Part d)

Using the given concentration and volume of HCl: \[ \text{Moles of HCl} = 0.208 \mathrm{~M} \times 0.0427 \mathrm{~L} = 0.00889 \mathrm{~moles} \]

Step 14: Calculate Moles of \(Ca(OH)_2\) (Part d)

Using the stoichiometry from Step 12: \[ \text{Moles of } Ca(OH)_2 = \frac{0.00889}{2} = 0.004445 \mathrm{~moles} \]

Step 15: Calculate Mass of \(Ca(OH)_2\) in Solution (Part d)

Find the mass of \(Ca(OH)_2\): \[ \text{Molar mass of } Ca(OH)_2 = 40 + 2(17) = 74 \mathrm{~g/mol} \] \[ \text{Mass} = 0.004445 \times 74 = 0.329 \mathrm{~g} \]

Key Concepts

Molarity

Molarity is a key concept when understanding how chemical reactions occur in solutions. It measures the concentration of a solute in a solution. Molarity (\(M\)) is expressed as the number of moles of solute per liter of solution.
For example, when you see a notation like \(0.120 \, M\) HCl, it means that one liter of the solution contains 0.120 moles of hydrochloric acid. Calculating molarity is crucial for determining how much of a reactant is required or produced in a chemical reaction.
To find molarity, use the formula:\[ M = \frac{\text{moles of solute}}{\text{liters of solution}} \]Using molarity allows chemists to accurately predict and manipulate reactions by ensuring the correct quantities of each substance are combined. It's like having the exact ingredients needed in a recipe to ensure the dish turns out right.

Stoichiometry

Stoichiometry is the practice of using balanced chemical equations to calculate the quantities of reactants and products in a chemical reaction. It is an essential tool for understanding how reactants interact to form products.
Stoichiometry helps predict how much of each reactant is required to complete a reaction without any excess. This involves ratios derived from the coefficients in a balanced chemical equation.
For instance, in the balanced equation \(2 \, \mathrm{HCl} + \mathrm{Ba(OH)}_2 \rightarrow 2 \, \mathrm{H}_2\mathrm{O} + \mathrm{BaCl}_2\), the coefficients indicate that two moles of hydrochloric acid react with one mole of barium hydroxide. Knowing these ratios allows chemists to compute the necessary amounts to avoid wastage or the need for additional reactant after the reaction starts.Stoichiometry is precise and provides a clear path from reactants to products, ensuring that the reaction proceeds as intended.

Balanced Chemical Equation

A balanced chemical equation is fundamental to studying chemical reactions because it shows the exact proportion of each substance involved in the reaction. This balance ensures that the law of conservation of mass is upheld, meaning the mass of reactants equals the mass of products.
Every balanced equation indicates the correct ratio in which reactants combine and the products they form. For instance, the balanced equation for reaction (a) is:\[ 2 \mathrm{HCl} + \mathrm{Ba(OH)}_2 \rightarrow 2 \mathrm{H}_2\mathrm{O} + \mathrm{BaCl}_2 \]This equation reveals that two moles of HCl are required for each mole of \(Ba(OH)_2\). Understanding the equation informs the quantities needed, maintaining the balance necessary for a complete reaction.
Each side of the equation signifies how the number of atoms of each element remains unchanged and contributes to ensuring the correct proportions in all calculations.

Solution Concentration

Solution concentration is vital in understanding how much solute is within a given amount of solvent. It is key to preparing solutions with precise properties for desired chemical reactions.
Concentration is often expressed in molarity, but it can also use other measures such as percent composition by volume or mass. The concentration informs us about the solution's strength, affecting how reactants interact and the overall reaction outcomes.
By knowing the concentration, such as \(0.101 \, M\) of \(Ba(OH)_2\), we understand how the solution will behave in a reaction and what quantity is necessary for neutralizing an acid effectively. For example, in a neutralization reaction, concentrations are adjusted to ensure all acid has reacted with the base, leaving neither in excess.Concentration allows for precise control over reactions, ensuring consistency and efficiency in chemical processes.