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
anor277
Apr 27, 2017

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.

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