Thank you for visiting A 38 2 mL sample of ammonia requires 21 2 mL of a 0 358 M solution of hydrochloric acid to reach equivalence point What. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!
Answer :
The concentration of the ammonia sample is 0.1985 M.
To find the concentration of the ammonia sample, we can use the concept of stoichiometry and the balanced chemical equation between ammonia (NH_3) and hydrochloric acid (HCl):
[tex]NH_3[/tex] + HCl > [tex]NH_4Cl[/tex]
Given that 38.2 mL of ammonia requires 21.2 mL of a 0.358 M hydrochloric acid solution to reach the equivalence point, we can set up a stoichiometry calculation to determine the concentration of the ammonia sample.
Step 1: Convert the volume of hydrochloric acid solution to moles:
Moles of HCl = volume of HCl solution (in L) × concentration of HCl (in M)
Moles of HCl = 21.2 mL ÷ 1000 mL/L × 0.358 M
Moles of HCl = 0.0075916 mol
Step 2: Use the balanced equation to find the moles of ammonia:
From the balanced equation, we know that 1 mole of [tex]NH_3[/tex] reacts with 1 mole of HCl. Therefore, the moles of ammonia are equal to the moles of hydrochloric acid.
Moles of [tex]NH_3[/tex] = Moles of HCl = 0.0075916 mol
Step 3: Convert the moles of ammonia to concentration:
Concentration of [tex]NH_3[/tex] = moles of [tex]NH_3[/tex] ÷ volume of [tex]NH_3[/tex] (in L)
Concentration of [tex]NH_3[/tex] = 0.0075916 mol ÷ 0.0382 L
Concentration of [tex]NH_3[/tex] = 0.1985 M
Therefore, the concentration of the ammonia sample is 0.1985 M.
To know more about stoichiometery refer here: https://brainly.com/question/13915354#
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