CaCrO4 is a slightly soluble solid. In the lab, you find that addition of 987.65
ID: 487003 • Letter: C
Question
CaCrO4 is a slightly soluble solid. In the lab, you find that addition of 987.65 µL of 1 M H2CrO4 to 4.567 mL of 0.3 M Ca(NO3)2 results in formation of a persistent precipitate. Addition of water (1.234 mL) just dissolves the precipitate.
What is the experimental value of Ksp for CaCrO4 according to the information above? Provide your response to 4 digits after the decimal.
A following way is WRONG.
Ca(NO, (aq) Cro. (aq e Cacro. (s) (aq) moles of H Cro added molarityxvolume volume 323.10 323.10x105L. Molarity 1M moles of H Crou added J 1x32 3.10x10 323-10x10 Moles of Cal NO. olarityx volume molarity 0.3M volume 5.026ml 5.026x10 L Moles of Ca(NO,), -0.3x5.026x10 5078x10 moles ratio and also thelimiting reagentisH2Cro. reactanats react in 1:1 molesof Cacro, formed moles of H Cro4 reacted 32 3.10x10 moles Since, imolesof Cro, giveslmoleof CaCro. Total volumeof solution 323.10x10 L+5.026x10 L+8.125x103L 0.0 1347L. solubility 323.10x10 moles of CaCro, per 0.01 347L 32 3.10x10 moles 0.01 347L. J0.023979molesL. dissociation of CaCro is as follows: CaCro, (s (s) (s) 0.023979 5.7499 x10Explanation / Answer
First we have to find wheather the reactant is present in excess
no. moles of CrO42- = 0.9876 mL x 1 M => 0.9876 mmol
no of moles of Ca2+ = 4.567 mL x 0.3 M => 1.3701 mmol
so Ca2+ is present in excess
the excess moles is
1.3701 - 0.9876 = 0.3825 mmol
The concentration of CaCrO4 formed depends upon the limiting reagent. Here limiting reagent is H2CrO4
So [CaCrO4] = 0.9876 mmol
After adding water(1.234 mL) total volume of the solution is = 0.9876 mL + 4.567 mL + 1.234 mL
= 6.7886 mL
The solublity of CaCrO4 is
= 0.9876mmol / 6.7886 mL
= 0.1454 mol/L
Thus concentration of Ca2+ after adding water is = 0.3825 / (0.9876 +4.567 + 1.234) => 0.0563 M
CaCrO4 -----> Ca2+ + CrO42-
Ksp = ( [Ca2+] + [excess concentration of Ca2+ from the reactant] ) [CrO42- ]
Ksp = (0.1454 + 0.0563) x (0.1454)
Ksp = 0.02932718