How much 2 M HBr is needed to neutralize 380 mL of 0.1 M NH4OH?
This is an acid-base neutralization reaction as well as a titration problem.
The key to problems of this type is to first write a balanced chemical equation showing the reactants and products.
HBr + NH4OH --> NH4Br + HOH
Note that in this reaction it takes one mole of HBr to react with one mole of NH4OH producing one mole of NH4Br and one mole of water (HOH)
The second key is to recognize that if you multiply the volume times the molarity the product is the number of moles of that material.
In short, mL * M of HBr = mL x M of NH4OH
Use this equation, substitute in what you know and solve for the missing variable.
M of HBr = 2.0
M of NH4OH = 0.1
mL of NH4OH = 380
Unknown is mL of HBr
2.0 * mL HBr = 0.1 * 380
mL HBr = 38/2 = 19 mL of HBr