Respuesta :
The equal volumes of 0.0300 m nh3 and 0.0150 m hcl are mixed together the concentrations of nh3 and nh4 in the reaction are 0.0075 M .
Suppose the every identical quantity is V.
- No. of moles of NH3 = 0.030 x V = 0.030V
- No. of moles of HCl = 0.half x V = 0.015V
- The 0.015V moles of NH3 are neutralized with 0.015V moles of HCl
- = No. of moles of NH4+ = 0.015V moles
- Total quantity = V + V = 2V
- Concentration of ammonium, [NH4+] = No. of moles/ quantity
- = 0.015V/2V = 0.0075 M (Ans)
- Non-reacting no. of moles of NH3 = (Total No. of moles of NH3) – (No. of moles of NH4+)
- = 0.030V - 0.015V moles = 0.015V moles
- Concentration of ammonia, [NH3] = No. of moles/ quantity
- = 0.015V/2V = 0.0075 M.
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