What volume of #HCl(g)# is required for equivalence to a #32*mL# volume of #AgNO_3(aq)# at #0.08*mol*L^-1# concentration?

1 Answer
Apr 27, 2017

Answer:

We need an equivalent molar quantity of #HCl(g)#......I gets under #60*cm^3# or under #60*mL#.......................

Explanation:

#"Moles of Ag"""^+=32xx10^-3Lxx0.08*mol*L^-1=2.56xx10^-3*mol#.

And thus we need an equivalent molar quantity of #Cl^-#, and thus an equivalent molar quantity of #HCl(g)#, which is presumed to dissolve quantitatively in water to give #HCl(aq)#.

So we use the Ideal Gas Equation,

#V=(nRT)/P=(2.56xx10^-3*molxx0.0821*(L*atm)/(K*mol)xx273*K)/(1*atm)#

Note that I have used #1*atm# to represent a pressure of #100*kPa#, which is near enuff.

We get #V=0.057*L=57.4*cm^3#.