A sparingly soluble salt gets precipitat...

A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution `(Q_(sp))` becomes greater than its solubility product. If the solubility of `BaSO_(4)` in water is `8 xx 10^(-4)` mol `dm^(-3)`. Calculate its solubility in `0.01` mol `dm^(-3)` of `H_(2)SO_(4)`. Express your answer in scientific notation `x xx 10^(-y)`. Write the value of y.

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The correct Answer is:

`0005`

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A sparingly soluble salt gets precipitated only when the prudct of concentration of its ions in th e solution (Q_(sp)) becomes greater than its solubility product. If solubility of BaSO_(4) in water is 8xx10^(-4)" mol dm"^(-3) . Calculater its solubility in 0.01" mol dm"^(-3) " of " H_(2)SO_(4) .

Higher the amount of acid or base used to product a deinite change of pH in a buffer solution, higher will be its buffe capacity. Buffer capcity of solution is maximum under the following conditions [Salt] = [Acid](in acid buffer) [Salt] = [Base] (in base capacity] pH of buffer solution lies inthe range given below pH = pH_(a) +-1 In other words,any buffer solution can be used as buffer up to two pH units only, depending upon the value of pK_(a) or pK_(b) . A buffer is said to be efficient when pH = pK_(a) or pOH = pK_(b) A sparingly soluble salt gets precipitated only when the product of concentration of its ions in the solution (Q_(sp)) becomes greater than its solubility product. If the solubility of BaSO_(4) in water is 8 xx 10^(-4) "mol" dm^(-3) calculate its solubility in 0.01 "mol" dm^(-3) of H_(2)SO_(4)

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