# 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?

Apr 27, 2017

We need an equivalent molar quantity of $H C l \left(g\right)$......I gets under $60 \cdot c {m}^{3}$ or under $60 \cdot m L$.......................

#### Explanation:

${\text{Moles of Ag}}^{+} = 32 \times {10}^{-} 3 L \times 0.08 \cdot m o l \cdot {L}^{-} 1 = 2.56 \times {10}^{-} 3 \cdot m o l$.

And thus we need an equivalent molar quantity of $C {l}^{-}$, and thus an equivalent molar quantity of $H C l \left(g\right)$, which is presumed to dissolve quantitatively in water to give $H C l \left(a q\right)$.

So we use the Ideal Gas Equation,

$V = \frac{n R T}{P} = \frac{2.56 \times {10}^{-} 3 \cdot m o l \times 0.0821 \cdot \frac{L \cdot a t m}{K \cdot m o l} \times 273 \cdot K}{1 \cdot a t m}$

Note that I have used $1 \cdot a t m$ to represent a pressure of $100 \cdot k P a$, which is near enuff.

We get $V = 0.057 \cdot L = 57.4 \cdot c {m}^{3}$.