Question

How many milliliters of \(2.00 \mathrm{M} \mathrm{HCl}\) must be added to neutralize the following solutions? (a) A mixture of \(0.160 \mathrm{MHNO}_{3}(100.0 \mathrm{~mL})\) and \(0.100 \mathrm{M} \mathrm{KOH}\) \((400.0 \mathrm{~mL})\) (b) A mixture of \(0.120 \mathrm{M} \mathrm{NaOH}(350.0 \mathrm{~mL})\) and \(0.190 \mathrm{M} \mathrm{HBr}\) \((150.0 \mathrm{~mL})\)

Short answer

(a) 12.0 mL; (b) 6.75 mL.

Step-by-step solution

Determine moles of each reactant (a)

Calculate the moles of nitric acid \( (\mathrm{HNO}_3) \) and moles of potassium hydroxide \( (\mathrm{KOH}) \) in the given volumes. \[moles\, \mathrm{HNO}_3 = 0.160\, \text{M} \times 0.100\, \text{L} = 0.0160\, \text{moles} \moles\, \mathrm{KOH} = 0.100\, \text{M} \times 0.400\, \text{L} = 0.0400\, \text{moles}\]

Determine net moles to be neutralized (a)

Since \(\mathrm{HNO}_3\) and \(\mathrm{KOH}\) react in a 1:1 ratio, calculate the excess moles. \[\text{Excess } \mathrm{KOH} = 0.0400\, \text{moles} - 0.0160\, \text{moles} = 0.0240\, \text{moles}\]

Calculate volume of \(2.00 \mathrm{M} \mathrm{HCl}\) needed (a)

Calculate the volume of \(\mathrm{HCl}\) required to neutralize the excess \(\mathrm{KOH}\):\[V = \frac{0.0240\, \text{moles}}{2.00\, \text{M}} = 0.0120\, \text{L} = 12.0\, \text{mL}\]

Determine moles of each reactant (b)

Calculate the moles of sodium hydroxide \( (\mathrm{NaOH}) \) and moles of hydrobromic acid \( (\mathrm{HBr}) \) in the given volumes.\[moles\, \mathrm{NaOH} = 0.120\, \text{M} \times 0.350\, \text{L} = 0.0420\, \text{moles} \moles\, \mathrm{HBr} = 0.190\, \text{M} \times 0.150\, \text{L} = 0.0285\, \text{moles}\]

Determine net moles to be neutralized (b)

Calculate the excess moles after neutralization of \(\mathrm{HBr}\) and \(\mathrm{NaOH}\) (1:1 reaction ratio).\[\text{Excess } \mathrm{NaOH} = 0.0420\, \text{moles} - 0.0285\, \text{moles} = 0.0135\, \text{moles}\]

Calculate volume of \(2.00 \mathrm{M} \mathrm{HCl}\) needed (b)

Calculate the volume of \(\mathrm{HCl}\) required to neutralize the excess \(\mathrm{NaOH}\):\[V = \frac{0.0135\, \text{moles}}{2.00\, \text{M}} = 0.00675\, \text{L} = 6.75\, \text{mL}\]

Key concepts

Molarity

Molarity is key when dealing with solutions in chemistry because it measures concentration. Specifically, it tells us how many moles of a solute are present in one liter of solution. This concentration, noted as "M," is expressed in moles per liter (mol/L).

In the original exercise, understanding molarity was crucial because we needed to find out how many moles of reactants were present to neutralize a solution.

• For example, in part (a), the solution of nitric acid (HNO₃) has a molarity of 0.160 M, which informs us that in every liter of this solution, there are 0.160 moles of HNO₃.
• We used this information alongside the given volume to calculate moles: Molarity imes Volume = Moles.

By applying this formula, we could find out exactly how many moles of acids or bases were present, allowing us to proceed with the stoichiometric calculations needed for neutralization.

Stoichiometry

Stoichiometry is the mathematical backbone of chemistry, describing the quantitative relationships in a chemical reaction. It essentially answers the question, "How much of each reactant is needed to reach the desired product?"

In acid-base neutralization problems like the one from the original exercise, stoichiometry helps us determine the amount of acid or base needed to fully react and neutralize a solution.

• First, we determine the number of moles of each reactant using calculated molarity and volume (Step 1 in our solution).
• Then, stoichiometry guides us in balancing the equation based on mole ratios, which tells us which reactant is in excess and can even predict the amounts of other products formed. In our case, both reactions had a 1:1 ratio between acid and base, simplifying our calculations.

This method ensures no excess reactant remains unreacted and the solution is properly neutralized.

Reaction Ratios

Reaction ratios refer to the amounts of reactants that participate in a chemical reaction. They are usually derived from the coefficients of a balanced chemical equation.

In acid-base neutralization reactions, knowing the reaction ratios is vital for calculating how much of each substance is needed for complete reaction. This helps in predicting when a solution will be fully neutralized.

• For example, in both scenarios from the exercise, the acids and bases reacted in a 1:1 ratio. This means that one mole of acid neutralizes one mole of base, and vice versa.
• This simplifies calculations since we only need to find out which reactant is in excess, and by how much, to add the exact volume of the neutralizing agent.

Understanding these ratios allows chemists to precisely control chemical reactions and make necessary adjustments efficiently, providing a complete solution to neutralization problems.