Question

What volume of 0.0850 M HCl is required to titrate 25.00 mL of a 0.100 M NH₃ solution to the equivalence point??

Please show work so I can understand -- thank you.

Answer 1

M1 = 0.085 M

V = 25 ml

M2 = 0.1 NH3

V2 = ?

this one is pretty easy!

The equivalence point is when the amount of moles of acid is equal to that of the base...

Acid HCL

Base NH3

1 mol of HCl will neutralize 1 mol of NH3

Calculate moles!

moles of NH3 = ?

M = mol/1L (this is the formula of molarity, a type of concentration)

0.1 M = mol NH3 / 25ml/1000ml/L (I just converted the 25 ml to liters, since the formula needs to be expressed in liters)

0.1M * (25/1000) = mol NH3

mol NH3 = 0.0025

Then we need 0.0025 mol of HCL

lets calculate that

M = mol Hcl / Vol HCl (in liters)

0.085 M = mol HCl / Vol HCl

mol HCl in equivalence point = mol of NH3 in equivalence point

mol HCl = 0.0025 mol

0.0085 M = 0.0025 mol / Vol HCl

Vol Hcl = 0.0025/0.0085 = 0.30 L

Vol HCl acid in mL = 0.30 L or 300 mL