3. (a) Calculate the volume of 0.5 M BaCl2, required for complete precipitation

Question

3.

(a) Calculate the volume of 0.5 M BaCl2, required for complete precipitation of 2.0 mL of the unknown solution (K2SO4) if its concentration is 0.10 M.

(b) Calculate the volume of reagent required for complete precipitation of 2.0 mL of the unknown solution (K2SO4) if its concentration is 1.0 M.

(c) If it turns out that about 1.5 mL of precipitating reagent is required for complete precipitation of 2.0 mL of unknown solution (K2SO4), what is its approximate concentration?

(d) Assuming the unknown concentration calculated in (c), how much unknown would you need to use in a precipitation to determine the concentration to four significant figures? What minimum volume of precipitating reagent would be required? What mass of precipitate would form? (use 10.00ml of unknown (K2SO4) and 10.00ml BaCl2 for 4 sigfigs, what is the ratio of the two?)

Explanation / Answer

BaCl2 + K2SO4 -----> 2KCl + BaSO4

1) As per the balanced reaction, BaCl2 & K2SO4 reacts in the molar ratio of 1:1

Thus, moles of BaCl2 required = moles of K2SO4 reacting

or, 0.5*V = 0.1*2

or, Volume of BaCl2 solution required = V = 0.4 ml

2) moles of BaCl2 required = moles of K2SO4 reacting

or, 0.5*V = 1*2

or, Volume of BaCl2 solution required = V = 4 ml

3) moles of BaCl2 required = moles of K2SO4 reacting

or, 0.5*1.5 = M*2

or, Molarity of K2SO4 solution required = M = 0.375 M

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