314975
The solubility of \(\mathrm{BaSO}_{4}\) in water is \(2.42 \times 10^{-3} \mathrm{gL}^{-1}\) at \(298 \mathrm{~K}\). The value of its solubility product \(\left( {{{\text{K}}_{{\text{sp}}}}} \right)\) will be
\(\mathrm{K}_{\mathrm{sp}}=4.9 \times 10^{-13}, \mathrm{~S}=\) ? \(\mathrm{AgBr}\) is \(\mathrm{AB}\) type salt. \(\text { So, } \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2} \quad \therefore \mathrm{S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}\) \(\therefore \mathrm{S}=\sqrt{49 \times 10^{-14}}=7.0 \times 10^{-7} \mathrm{moldm}^{-3}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
314977
The solubility of sparingly soluble salt \(\mathrm{AB}_{2}\) is \(1.0 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\). What is it's solubility product?
1 \(2 \times 10^{-12}\)
2 \(4 \times 10^{-8}\)
3 \(4 \times 10^{-12}\)
4 \(2 \times 10^{-8}\)
Explanation:
\(\mathrm{S}=1.0 \times 10^{-4} \mathrm{moldm}^{-3}, \mathrm{~K}_{\mathrm{sp}}=\) ? for \(\mathrm{AB}_{2}\) type salts, \(\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^{3}\) \(=4 \times\left(1.0 \times 10^{-4}\right)^{3}=4 \times 10^{-12}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
314978
Solubility of \(\mathrm{AgCl}\) is \(7.2 \times 10^{-7} \mathrm{moldm}^{-3}\). What is it's solubility product?
1 \(3.6 \times 10^{-13}\)
2 \(7.2 \times 10^{-14}\)
3 \(2.59 \times 10^{-14}\)
4 \(5.18 \times 10^{-13}\)
Explanation:
For \(\mathrm{AgCl}, \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2}\) (AB type of salt) \(\mathrm{S}=7.2 \times 10^{-7} \mathrm{moldm}^{-3}\) \(\mathrm{K}_{\mathrm{sp}}=\left(7.2 \times 10^{-7}\right)^{2}=5.18 \times 10^{-13}\)
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CHXI07:EQUILIBRIUM
314975
The solubility of \(\mathrm{BaSO}_{4}\) in water is \(2.42 \times 10^{-3} \mathrm{gL}^{-1}\) at \(298 \mathrm{~K}\). The value of its solubility product \(\left( {{{\text{K}}_{{\text{sp}}}}} \right)\) will be
\(\mathrm{K}_{\mathrm{sp}}=4.9 \times 10^{-13}, \mathrm{~S}=\) ? \(\mathrm{AgBr}\) is \(\mathrm{AB}\) type salt. \(\text { So, } \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2} \quad \therefore \mathrm{S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}\) \(\therefore \mathrm{S}=\sqrt{49 \times 10^{-14}}=7.0 \times 10^{-7} \mathrm{moldm}^{-3}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
314977
The solubility of sparingly soluble salt \(\mathrm{AB}_{2}\) is \(1.0 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\). What is it's solubility product?
1 \(2 \times 10^{-12}\)
2 \(4 \times 10^{-8}\)
3 \(4 \times 10^{-12}\)
4 \(2 \times 10^{-8}\)
Explanation:
\(\mathrm{S}=1.0 \times 10^{-4} \mathrm{moldm}^{-3}, \mathrm{~K}_{\mathrm{sp}}=\) ? for \(\mathrm{AB}_{2}\) type salts, \(\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^{3}\) \(=4 \times\left(1.0 \times 10^{-4}\right)^{3}=4 \times 10^{-12}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
314978
Solubility of \(\mathrm{AgCl}\) is \(7.2 \times 10^{-7} \mathrm{moldm}^{-3}\). What is it's solubility product?
1 \(3.6 \times 10^{-13}\)
2 \(7.2 \times 10^{-14}\)
3 \(2.59 \times 10^{-14}\)
4 \(5.18 \times 10^{-13}\)
Explanation:
For \(\mathrm{AgCl}, \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2}\) (AB type of salt) \(\mathrm{S}=7.2 \times 10^{-7} \mathrm{moldm}^{-3}\) \(\mathrm{K}_{\mathrm{sp}}=\left(7.2 \times 10^{-7}\right)^{2}=5.18 \times 10^{-13}\)
314975
The solubility of \(\mathrm{BaSO}_{4}\) in water is \(2.42 \times 10^{-3} \mathrm{gL}^{-1}\) at \(298 \mathrm{~K}\). The value of its solubility product \(\left( {{{\text{K}}_{{\text{sp}}}}} \right)\) will be
\(\mathrm{K}_{\mathrm{sp}}=4.9 \times 10^{-13}, \mathrm{~S}=\) ? \(\mathrm{AgBr}\) is \(\mathrm{AB}\) type salt. \(\text { So, } \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2} \quad \therefore \mathrm{S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}\) \(\therefore \mathrm{S}=\sqrt{49 \times 10^{-14}}=7.0 \times 10^{-7} \mathrm{moldm}^{-3}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
314977
The solubility of sparingly soluble salt \(\mathrm{AB}_{2}\) is \(1.0 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\). What is it's solubility product?
1 \(2 \times 10^{-12}\)
2 \(4 \times 10^{-8}\)
3 \(4 \times 10^{-12}\)
4 \(2 \times 10^{-8}\)
Explanation:
\(\mathrm{S}=1.0 \times 10^{-4} \mathrm{moldm}^{-3}, \mathrm{~K}_{\mathrm{sp}}=\) ? for \(\mathrm{AB}_{2}\) type salts, \(\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^{3}\) \(=4 \times\left(1.0 \times 10^{-4}\right)^{3}=4 \times 10^{-12}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
314978
Solubility of \(\mathrm{AgCl}\) is \(7.2 \times 10^{-7} \mathrm{moldm}^{-3}\). What is it's solubility product?
1 \(3.6 \times 10^{-13}\)
2 \(7.2 \times 10^{-14}\)
3 \(2.59 \times 10^{-14}\)
4 \(5.18 \times 10^{-13}\)
Explanation:
For \(\mathrm{AgCl}, \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2}\) (AB type of salt) \(\mathrm{S}=7.2 \times 10^{-7} \mathrm{moldm}^{-3}\) \(\mathrm{K}_{\mathrm{sp}}=\left(7.2 \times 10^{-7}\right)^{2}=5.18 \times 10^{-13}\)
314975
The solubility of \(\mathrm{BaSO}_{4}\) in water is \(2.42 \times 10^{-3} \mathrm{gL}^{-1}\) at \(298 \mathrm{~K}\). The value of its solubility product \(\left( {{{\text{K}}_{{\text{sp}}}}} \right)\) will be
\(\mathrm{K}_{\mathrm{sp}}=4.9 \times 10^{-13}, \mathrm{~S}=\) ? \(\mathrm{AgBr}\) is \(\mathrm{AB}\) type salt. \(\text { So, } \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2} \quad \therefore \mathrm{S}=\sqrt{\mathrm{K}_{\mathrm{sp}}}\) \(\therefore \mathrm{S}=\sqrt{49 \times 10^{-14}}=7.0 \times 10^{-7} \mathrm{moldm}^{-3}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
314977
The solubility of sparingly soluble salt \(\mathrm{AB}_{2}\) is \(1.0 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}\). What is it's solubility product?
1 \(2 \times 10^{-12}\)
2 \(4 \times 10^{-8}\)
3 \(4 \times 10^{-12}\)
4 \(2 \times 10^{-8}\)
Explanation:
\(\mathrm{S}=1.0 \times 10^{-4} \mathrm{moldm}^{-3}, \mathrm{~K}_{\mathrm{sp}}=\) ? for \(\mathrm{AB}_{2}\) type salts, \(\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^{3}\) \(=4 \times\left(1.0 \times 10^{-4}\right)^{3}=4 \times 10^{-12}\)
MHTCET - 2021
CHXI07:EQUILIBRIUM
314978
Solubility of \(\mathrm{AgCl}\) is \(7.2 \times 10^{-7} \mathrm{moldm}^{-3}\). What is it's solubility product?
1 \(3.6 \times 10^{-13}\)
2 \(7.2 \times 10^{-14}\)
3 \(2.59 \times 10^{-14}\)
4 \(5.18 \times 10^{-13}\)
Explanation:
For \(\mathrm{AgCl}, \mathrm{K}_{\mathrm{sp}}=\mathrm{S}^{2}\) (AB type of salt) \(\mathrm{S}=7.2 \times 10^{-7} \mathrm{moldm}^{-3}\) \(\mathrm{K}_{\mathrm{sp}}=\left(7.2 \times 10^{-7}\right)^{2}=5.18 \times 10^{-13}\)