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leva [86]
2 years ago
12

What is ksp for agi, if its molar solubility is 9.1 × 10−9 mol/l?

Chemistry
1 answer:
LekaFEV [45]2 years ago
4 0
When the salt AgI dissolves, it dissociates as follows;
AgI --> Ag⁺ + I⁻
molar solubility of salt is the amount of salt that can be dissolved in 1 L of solution
since the ions dissociated are in 1:1 molar ratio, the molar solubility of the ions are equivalent to the molar solubility of the salt.
ksp is the solubility product constant of the salt
ksp = [Ag⁺][I⁻]
ksp = (9.1 x 10⁻⁹ mol/L)²
ksp = 8.28 x 10⁻¹⁷
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Hello,

The suitable differential equation for this case is:

\frac{dV}{dt}=10\frac{cm^3}{s}

As we're looking for the change in height with respect to the time, we need a relationship to achieve such as:

\frac{dh}{dt} = ?*\frac{dV}{dt}

Of course, ?=\frac{dh}{dV}.

Now, since the volume of a cone is V=\pi r^2h/3 and the ratio r/h=15/20=3/4 or r=3/4h, the volume becomes:

V=\pi (\frac{3}{4} h)^2h/3= \frac{3}{16}\pi h^3

We proceed to its differentiation:

\frac{dV}{dh} =\frac{9}{16} \pi h^2\\\frac{dh}{dV} =\frac{16}{9 \pi h^2}

Then, we compute \frac{dh}{dt}

\frac{dh}{dt} = \frac{16}{9 \pi h^2}*\frac{dV}{dt}\\\frac{dh}{dt} = \frac{16}{9\pi h^2}*10\frac{cm^3}{s} =\frac{160}{9 \pi h^2}

Finally, at h=2:

\frac{dh}{dt}_{h=2cm} =\frac{160}{9\pi 2^2}\\\frac{dh}{dt}_{h=2cm} =\frac{40}{9\pi}\frac{cm}{s}

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