Given, K_{p} for the reaction i.e., H_{2} (g) + Br_{2} (g)** ⇒ **2HBr (g) is

\( 1.6 \times 10^{5}\)

Therefore, for the reaction 2HBr (g)** **

**⇒ **H_{2} (g) + Br_{2} (g) the equilibrium constant will be,

\( K’p= \dfrac{1}{K_{p}}\\ \\ =\dfrac{1}{1.6\times 10^{5}}\\ \\ =6.25\times 10^{-6}\)

Now, let p be the pressure of both H_{2} and Br_{2 }at equilibrium.

2HBr (g) **⇌** H_{2} (g) + Br_{2} (g)

Initial conc. 10 0 0

At equilibrium 10-2p p p

Now, we can write,

\( \dfrac{p_{HBr}\times p_{2}}{p^{2}_{HBr}}=K^{‘}_{p}\\ \\\)

\( \dfrac{p\times p}{(10-2p)^{2}}=6.25\times 10^{-6}\\ \\ \dfrac{p}{10-2p}=2.5\times 10^{-3}\\ \\ \)

\( p=2.5\times 10^{-2}-(5.0\times 10^{-3})p\\ \\ p+(5.0\times 10^{-3})\)

\( p=2.5\times 10^{-2}\\ \\ (1005\times 10^{-3})\)

\( =2.5\times 10^{-2}\\ \\\)

\( p=2.49\times 10^{-2}bar\)

\( =2.5\times 10^{-2}bar (approximately)\)

Therefore, at equilibrium,

\( [H_2]=[Br_2]= 2.49\times 10^{-2}bar[HBr] \)

\( = 10-2\times (2.49\times 10^{-2})bar\\ \\ \)

\( =9.95 bar=10 bar(approximately) \)

The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10^{–19} M. If 10 mL of this is added to 5 mL of 0.04M solution of the following: FeSO_{4}, MnCl_{2}, ZnCl_{2} and CdCl_{2} . in which of these solutions precipitation will take place?

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, K_{sp} is 9.1 × 10^{–6}).

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, Ksp = 6.3 × 10^{–18}).

The ionization constant of benzoic acid is 6.46 × 10^{–5} and Ksp for silver benzoate is 2.5 × 10^{–13}. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate K_{sp} = 7.4 × 10^{–8} ).