Asked by Emma
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...
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...
Answers
Answered by
DrBob222
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.
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.
Answered by
Damon
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
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
Answered by
pp
yamom