
The solubility product of $BaS{O_4}$ at $25 \circ C$ is $1.0 \times {10^{ - 9}}$ .What would be the concentration of ${H_2}S{O_4}$ required to precipitate $BaS{O_4}$ from a solution of $0.01M$ $B{a^{2 + }}$ ions.
Answer
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Hint: $BaS{O_4}$ is the chemical formula for Barium sulphate.It’s almost insoluble in water at room temperature and is used as oil well drilling fluid. ${H_2}S{O_4}$ is the chemical formula for Sulphuric acid which is a very strong mineral acid.
Complete answer:
For the above given question let us analyse the given compound Barium sulphate.We know that it is formed of two ions-Barium and sulphate.The chemical reaction for the same is given as
$BaS{O_4} \to B{a^{2 + }} + SO_4^{2 - }$ .In the above question it is given that the solubility product for this reaction at $25 \circ C$ is $1.0 \times {10^{ - 9}}$ ,i.e ${K_{sp}} = 1.0 \times {10^{ - 9}}$
But for the above reaction ${K_{sp}} = [B{a^{2 + }}][SO_4^{2 - }]$
Substituting the values which we have obtained from the above question.
${10^{ - 9}} = [B{a^{2 + }}][SO_4^{2 - }]$
But the value of $[B{a^{2 + }}]$ is given as 0.01.Therefore substituting the value in the equation we get.
${10^{ - 9}} = 0.01 \times [SO_4^{2 - }]$
Taking the constant on one side and others one one side we get,
$\frac{{{{10}^{ - 9}}}}{{0.01}} = [SO_4^{2 - }]$
Rearranging the terms and solving it we can get the concentration of sulphate ions.
$[SO_4^{2 - }] = {10^{ - 7}}$
Hence we now know that we need ${10^{ - 7}}M$ concentration of ${H_2}S{O_4}$ to precipitate $BaS{O_4}$ from a solution of $0.01M$ $B{a^{2 + }}$ ions.
Note:
The solubility product constant can be defined as the equilibrium constant for the dissolution of a solid substance into an aqueous solution.The symbol ${K_{sp}}$ is used to denote the solubility product of an given chemical equation.The general formula for it is given as,
${K_{sp}} = {[{A^ + }]^a}{[{B^ - }]^b}$ where ${A^ + }$ and ${B^ - }$ are the cations and anions respectively
For the above given equation,
A= $B{a^{2 + }}$ ,B= $SO_4^{2 - }$ and a and b are equal to 1.
Complete answer:
For the above given question let us analyse the given compound Barium sulphate.We know that it is formed of two ions-Barium and sulphate.The chemical reaction for the same is given as
$BaS{O_4} \to B{a^{2 + }} + SO_4^{2 - }$ .In the above question it is given that the solubility product for this reaction at $25 \circ C$ is $1.0 \times {10^{ - 9}}$ ,i.e ${K_{sp}} = 1.0 \times {10^{ - 9}}$
But for the above reaction ${K_{sp}} = [B{a^{2 + }}][SO_4^{2 - }]$
Substituting the values which we have obtained from the above question.
${10^{ - 9}} = [B{a^{2 + }}][SO_4^{2 - }]$
But the value of $[B{a^{2 + }}]$ is given as 0.01.Therefore substituting the value in the equation we get.
${10^{ - 9}} = 0.01 \times [SO_4^{2 - }]$
Taking the constant on one side and others one one side we get,
$\frac{{{{10}^{ - 9}}}}{{0.01}} = [SO_4^{2 - }]$
Rearranging the terms and solving it we can get the concentration of sulphate ions.
$[SO_4^{2 - }] = {10^{ - 7}}$
Hence we now know that we need ${10^{ - 7}}M$ concentration of ${H_2}S{O_4}$ to precipitate $BaS{O_4}$ from a solution of $0.01M$ $B{a^{2 + }}$ ions.
Note:
The solubility product constant can be defined as the equilibrium constant for the dissolution of a solid substance into an aqueous solution.The symbol ${K_{sp}}$ is used to denote the solubility product of an given chemical equation.The general formula for it is given as,
${K_{sp}} = {[{A^ + }]^a}{[{B^ - }]^b}$ where ${A^ + }$ and ${B^ - }$ are the cations and anions respectively
For the above given equation,
A= $B{a^{2 + }}$ ,B= $SO_4^{2 - }$ and a and b are equal to 1.
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