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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)

Respuesta :

W0lf93
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
sebassandin

Answer:

[tex]M_{NH_3}=2.98M[/tex]

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:

[tex]n_{NH_3}=n_{HCl}\\V_{NH_3}M_{NH_3}=V_{HCl}M_{HCl}\\M_{NH_3}=\frac{V_{HCl}M_{HCl}}{V_{NH_3}} \\M_{NH_3}=\frac{18.6mL*0.800M}{5.00mL}\\M_{NH_3}=2.98M[/tex]

Best regards.

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