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...

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  1. 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.

  2. 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

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