The reaction of hydrochloric acid (HCl) with ammonia (NH3) is described by the equation:
HCl + NH3 → NH4Cl
A student is titrating 50 mL of 0.32 M NH3 with 0.5 M HCl. How much hydrochloric acid must be added to react completely with the ammonia?
A. 6.4 mL
B. 16.0 mL
C. 32.0 mL
D. 50.0 mL
I think that you are supposed to use (H+)(MA*VA)= (MB*VB)(OH-) but I'm confused on the setup...
If you want to use a formula I would use MaVa = MbVb. I don't understand where the H and OH fits. Then
0.5*Va = 0.32*50 and solve for Va.
However, I think it's easier to understand how the titration works and go from there. Besides, the formula of MaVa = MbVb works ONLY for acids and bases that react 1:1. There are different formulas if the coefficients are not 1:1. I recommend this way.
Recognize that the end of the titration comes when the mols of acid = mols base.That is true for ANY titration. mols = M x L so mols NH3 = M x L = 0.32 x 0.05 = 0.016. The equation tells us that 1 mol NH3 will require 1 mol HCl; therefore, mols HCl = 0.016. Then mols HCl = M x L. You know mols = 0.016 and the problem tells you M = 0.5; therefore L = mols/M = 0.016/0.5 = ? L and you can convert to mL if you wish.
You need the same number of mols of HCl as you have of NH3
we have 50 mL of 0.32 M NH3
.32 mol / liter * 50*10^-3 L = 16 *10^-3 mol
we need that many mols of HCl
0.5 mol /liter * x liter = 16*10^-3 mol
x = 32 * 10^-3 L which is 32 mL
yamom
To determine how much hydrochloric acid (HCl) must be added to react completely with the ammonia (NH3), we can use the concept of stoichiometry and the balanced chemical equation provided. Here's how to set up the problem:
Step 1: Write the balanced chemical equation:
HCl + NH3 → NH4Cl
The equation shows that one mole of HCl reacts with one mole of NH3 to produce one mole of NH4Cl.
Step 2: Convert the given quantities to moles:
Given:
Volume of NH3 (VNH3) = 50 mL = 0.050 L
Molarity of NH3 (MNH3) = 0.32 M
Volume of HCl (VHCl) = ? (what we need to find)
Molarity of HCl (MHCl) = 0.5 M
To convert the volume to moles, we use the formula: Moles = Volume (L) × Concentration (M)
Moles of NH3 = VNH3 × MNH3 = 0.050 L × 0.32 M = 0.016 mol
Step 3: Use stoichiometry to determine the mole ratio:
From the balanced equation, we can see that the mole ratio between HCl and NH3 is 1:1. This means that 1 mole of HCl reacts with 1 mole of NH3. Therefore, the moles of HCl needed will be equal to the moles of NH3.
Step 4: Convert moles of HCl to the volume of HCl:
To find the volume of HCl needed, we can rearrange the formula: Moles = Volume (L) × Concentration (M) to solve for the volume.
VHCl = moles of HCl / MHCl
VHCl = 0.016 mol / 0.5 M = 0.032 L = 32 mL
Therefore, the correct answer is C. 32.0 mL.