First slide

Solubility product(KSP)

Question

Using the Gibb's energy change, $∆ G ° = + 63.3 k J,$ for the following reaction,
${Ag}_{2} {CO}_{3 (s)} ⇌ 2 {Ag}_{(a q)}^{+} + {CO}_{3 (a q)}^{2 -}$
the $K_{s p} of {Ag}_{2} {CO}_{3 (s)}$ in water at 25°C is
$(R = 8.314 J K^{- 1} {mol}^{- 1})$

Moderate

Solution

$Δ G ° = - 2.303 R T \log K_{s p}$
$63.3 \times 10^{3} J = - 2.303 \times 8.314 \times 298 \log K_{s p}$
$63.3 \times 10^{3} J = - 5705.84 \log K_{s p}$
$\log K_{s p} = - \frac{63.3 \times 10^{3}}{5705.84} = - 11.09$
$K_{s p} = antilog (- 11.09) = 8.128 \times 10^{- 12}$

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