A 100.0 ml solution of naoh reaches the equivalence point when 30.22 ml of a 0.0750 m solution of hcl is added from the burette.
The Correct Answer and Explanation is:
To solve this, let’s begin by using the concept of equivalence points in acid-base reactions. The equivalence point is reached when the number of moles of HCl (acid) added from the burette equals the number of moles of NaOH (base) in the solution. Given that we know the molarity (M) and volume (V) of HCl added, we can determine the moles of HCl and, subsequently, the concentration of NaOH.
Step-by-Step Solution
- Calculate the moles of HCl added:
- The volume of HCl solution added is 30.22 mL, or 0.03022 L (since (1 \, \text{mL} = 0.001 \, \text{L})).
- The molarity of HCl is 0.0750 M. Moles of ( \text{HCl} = M \times V ) [
\text{Moles of HCl} = 0.0750 \, \text{M} \times 0.03022 \, \text{L} = 0.0022665 \, \text{moles}
]
- Equivalence with NaOH:
- At the equivalence point, the moles of HCl added equal the moles of NaOH initially present in the 100.0 mL NaOH solution. Therefore, moles of ( \text{NaOH} = 0.0022665 \, \text{moles} ).
- Calculate the molarity of NaOH:
- The volume of the NaOH solution is 100.0 mL, or 0.1000 L. [
\text{Molarity of NaOH} = \frac{\text{moles of NaOH}}{\text{volume of NaOH solution in liters}} = \frac{0.0022665 \, \text{moles}}{0.1000 \, \text{L}} = 0.022665 \, \text{M}
] Rounding to three significant figures (based on the given data), the molarity of NaOH is 0.0227 M.
Explanation
In an acid-base titration, the equivalence point is where the acid completely neutralizes the base (or vice versa). Here, hydrochloric acid (HCl), a strong acid, neutralizes sodium hydroxide (NaOH), a strong base. At equivalence, the moles of HCl added from the burette are equal to the moles of NaOH present in the initial solution.
Knowing the molarity and volume of HCl, we used the relationship ( \text{Molarity (M)} = \frac{\text{moles}}{\text{Volume (L)}} ) to find the moles of HCl. Because these moles equal the moles of NaOH at equivalence, we then calculated the molarity of NaOH using the initial volume of its solution. Thus, the NaOH solution has a molarity of 0.0227 M. This process emphasizes stoichiometric balance in neutralization reactions and how concentration relationships help in finding unknown solution molarities.