Question

If 18.6 ml of 0.800 m hcl solution are needed to neutralize 5.00 ml of a household ammonia solution, what is the molar concentration of the ammonia? nh3(aq) + hcl(aq) →nh4cl(aq)

Answer 1

`M_(NH_3)=2.98M`

Step-by-step explanation:

Hello,

By using the following formula, one could compute the required molarity of the neutralized solution of ammonia since in the neutralization point the both acid's and base's moles must be equal as long as in the reaction a 1-to-1 mole relationship is present between them:

`n_(NH_3)=n_(HCl)\\V_(NH_3)M_(NH_3)=V_(HCl)M_(HCl)\\M_(NH_3)=(V_(HCl)M_(HCl))/(V_(NH_3)) \\M_(NH_3)=(18.6mL*0.800M)/(5.00mL)\\M_(NH_3)=2.98M`

Best regards.

Answer 2

Molarity of HCl = number of moles/ volume
Number of moles = Molarity of Hcl * volume
No of moles = 0.8 * 18.6 * 10^(-3) = 14.88 * 10^(-3)
From the equation one mole of HCl reacts with one mole of ammonia
Hence number of moles of ammonia = 14.88 * 10^(-3)
From question volume of ammonia = 5 * 10^(-3)
So it follows that Molarity of ammonia = number of moles of ammonia/ volume
Molarity = 14.88 * 10^(-3)/5 * 10^(-3) = 2.976 * 10^(-3 -(-3))
Molarity = 2.976