Which statement explains why 10.0 mL of a 0.50 M H2504(aq) solution exactly neutralizes 5.0
mL of a 2.0 M NaOH(aq) solution?
A)
The moles of H(aq) equal the moles of OH(aq).
B)
The moles of H2502(aq) equal the moles of NaOH(aq).
C)
The moles of H2SO4(aq) are greater than the moles of NaOH(aq).
D)
The moles of H(aq) are greater than the moles of OH(aq).
The Correct Answer and Explanation is :
The correct answer is:
A) The moles of H(^+)(aq) equal the moles of OH(^-)(aq).
Explanation:
In an acid-base neutralization reaction, the acid (H(^+)) reacts with the base (OH(^-)) to form water and a salt. The key to this problem is understanding the molarity and volume relationship between the acid and base and how they relate to the number of moles of ions involved in the reaction.
We are given the following:
- Volume of H(_2)SO(_4) solution: 10.0 mL (0.010 L)
- Concentration of H(_2)SO(_4): 0.50 M
- Volume of NaOH solution: 5.0 mL (0.005 L)
- Concentration of NaOH: 2.0 M
First, calculate the moles of H(_2)SO(_4) and NaOH using the formula:
[
\text{moles} = \text{Molarity} \times \text{Volume (in L)}
]
- Moles of H(_2)SO(_4):
[
\text{moles of H(_2)SO(_4)} = 0.50 \, \text{M} \times 0.010 \, \text{L} = 0.005 \, \text{moles of H(_2)SO(_4)}
] - Moles of NaOH:
[
\text{moles of NaOH} = 2.0 \, \text{M} \times 0.005 \, \text{L} = 0.010 \, \text{moles of NaOH}
]
The balanced equation for the neutralization of sulfuric acid (H(_2)SO(_4)) with sodium hydroxide (NaOH) is:
[
H_2SO_4(aq) + 2 \, NaOH(aq) \rightarrow Na_2SO_4(aq) + 2 \, H_2O(l)
]
This equation shows that 1 mole of H(_2)SO(_4) neutralizes 2 moles of NaOH. Since we have 0.005 moles of H(_2)SO(_4), we would need 0.010 moles of NaOH to neutralize it, which matches the given amount.
Thus, the moles of hydrogen ions (H(^+)) from H(_2)SO(_4) exactly equal the moles of hydroxide ions (OH(^-)) from NaOH, meaning A) The moles of H(^+) equal the moles of OH(^-).