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.
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
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