Question

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?

Answer 1

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`.