A 25 mL sample of $0.100 M HNO_3$ completely reacts with NaOH according to this equation:
$HNO_3(aq) + NaOH(aq) \rightarrow NaNO_3(aq) + H_2O(l)$
What volume of 0.0500 M NaOH solution is required to completely react with the $HNO_3$?
A. 2.50 mL NaOH
B. 5.00 mL NaOH
C. 25.0 mL NaOH
D. 50.0 mL NaOH
Answer (2)
Calculate the moles of H N O 3 using the formula: moles = Molarity × Volume (in Liters) , which gives 0.0025 mol.
Determine that the moles of N a O H required are equal to the moles of H N O 3 due to the 1:1 reaction ratio.
Calculate the volume of N a O H using the formula: Volume (in Liters) = Molarity moles , resulting in 0.05 L.
Convert the volume of N a O H to mL: 0.05 L × 1000 = 50 mL . The final answer is 50.0 mL .
Explanation
Problem Analysis We are given a problem where a 25 mL sample of 0.100 M H N O 3 completely reacts with NaOH. The balanced chemical equation is: H N O 3 ( a q ) + N a O H ( a q ) → N a N O 3 ( a q ) + H 2 O ( l ) We need to find the volume of 0.0500 M NaOH solution required to completely react with the H N O 3 .
Calculating Moles of HNO3 First, we need to calculate the number of moles of H N O 3 present in the 25 mL sample. We can use the formula: moles = Molarity × Volume (in Liters) Given: Volume of H N O 3 = 25 mL = 0.025 L Molarity of H N O 3 = 0.100 M moles of H N O 3 = 0.100 L mol × 0.025 L = 0.0025 mol
Determining Moles of NaOH Since the reaction between H N O 3 and N a O H is 1:1, it means that 1 mole of H N O 3 reacts with 1 mole of N a O H . Therefore, the number of moles of N a O H required is equal to the number of moles of H N O 3 .
moles of N a O H = 0.0025 mol
Calculating Volume of NaOH Now, we can calculate the volume of 0.0500 M N a O H solution required using the formula: Volume (in Liters) = Molarity moles Given: Molarity of N a O H = 0.0500 M Moles of N a O H = 0.0025 mol Volume of N a O H = 0.0500 L mol 0.0025 mol = 0.05 L
Final Volume of NaOH Finally, we convert the volume from Liters to mL by multiplying by 1000: Volume of N a O H = 0.05 L × 1000 L mL = 50 mL Therefore, 50.0 mL of 0.0500 M NaOH solution is required to completely react with the H N O 3 .
Examples
In acid-base titrations, like the reaction between H N O 3 and N a O H , determining the exact volume of a base needed to neutralize an acid is crucial. This principle is applied in various fields, such as environmental monitoring to measure the acidity of rainwater, in chemical laboratories to standardize solutions, and in pharmaceutical industries to ensure the correct formulation of medications. By accurately calculating the required volumes, we can control chemical reactions and maintain desired pH levels in different applications.
You need 50.0 mL of 0.0500 M NaOH to completely react with 25 mL of 0.100 M HNO3. This is calculated by first determining the moles of HNO3 and then using the 1:1 mole ratio to find the required volume of NaOH. After calculations, the final answer is 50.0 mL.
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